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REESE LIBRARY

UNIVERSITY OF CALIFORNIA.

Deceived , igo .

Accession No. 1)2307. Clots No. £4 777

ELEMENTS OF GRAPHIC STATICS.

THE

ELEMENTS OF GRAPHIC STATICS

for Students of 3Ensineering

L. M. HOSKINS

PROFESSOR OF APPLIED MECHANICS IN THE LELAND STANFORD
JUNIOR UNIVERSITY

REVISED EDITION

Xf|tTiiR4^

/V-OFTHE ^

gorft
THE MACMILLAN COMPANY

LONDON : MACMILLAN & CO., LTD.
1899

Set up and electrotyped August, 1892. New edition, revised,
printed from new plates January, 1899.

J. S. Gushing & Co. — Berwick & Smith
Norwood Mass. U.S.A.

PREFACE TO REVISED EDITION.

THE method of treatment adopted in this work is designed
to meet the needs of the beginner. To this end the endeavor
has been made to secure simplicity of presentation without
sacrifice of logical rigor.

In scope, the work has been planned with reference to the
requirements of students of engineering. This limits the de-
velopment of the general theory to such principles and methods
as are practically useful. It also excludes many applications
which, though leading to practical results, are likely to prove
useful and to save labor only in the hands of the expert in
graphical methods. Graphic Statics is treated as a branch of
Mechanics rather than of Geometry, and those beautiful devel-
opments whose chief interest is geometrical have not been
included.

Although graphical methods are especially useful to the
structural engineer, it is believed that students in all depart-
ments of engineering will find it profitable to become familiar
with the general theory of complanar forces from the graphical
side, as well as with the simpler applications to the determina-
tion of stresses in framed structures, and of centroids and mo-
ments of inertia of plane areas. The application to trusses
carrying moving loads is of less general interest, its practical
utility being limited mainly to the solution of problems in
bridge design. The methods developed in Chapter VII will
therefore be of practical value mainly to students giving special
attention to this branch of engineering.

In the present revised edition no change has been made in
general plan, and few changes in the treatment adopted, except

vi PREFACE TO THE REVISED EDITION.

in the portions relating to beams and trusses carrying moving
loads (Chapters VI and VII). These portions have been
wholly re-written. It is believed that a substantial improve-
ment has been made upon the methods hitherto used, par-
ticularly in the criterion for determining the position of a
given load-series which causes maximum stress in any member
of a truss. The improvement consists in generalization, which
is believed to be gained without sacrifice of simplicity. The
graphical method of applying the criterion in the case of
trusses with parallel chords has been fully treated by Pro-
fessor H. T. Eddy. The method here given applies without
the restriction to parallel chords. The algebraic statement of
the same criterion, as given in Art. 152, is also believed to
be a useful generalization of the methods heretofore used.
Whether the algebraic or the graphical treatment is preferred,
a method is useful in proportion to its generality, provided this
does not involve a loss of simplicity. There is a decided ad-
vantage in the use of a single general equation, applicable to
any member of any truss, instead of several particular equa-
tions, each applicable to a special member or to a special form
of truss.

STANFORD UNIVERSITY, CALIFORNIA,
November, 1898.

CONTENTS.

PART I. — GENERAL THEORY.
CHAPTER I. DEFINITIONS. — CONCURRENT FORCES.

PAGE

§ I. Preliminary Definitions . ' I

2. Composition of Concurrent Forces 5

3. Equilibrium of Concurrent Forces 6

4. Resolution of Concurrent Forces 8

CHAPTER II NON-CONCURRENT FORCES.

§ i. Composition of Non-concurrent Forces Acting on the Same Rigid

Body ii

2. Equilibrium of Non-concurrent Forces 18

3. Resolution into Non-concurrent Systems 28

4. Moments of Forces and of Couples 30

5. Graphic Determination of Moments 35

6. Summary of Conditions of Equilibrium 37

CHAPTER III. INTERNAL FORCES AND STRESSES.

§ i. External and Internal Forces 40

2. External and Internal Stresses 42

3. Determination of Internal Stresses 46

PART II. — STRESSES IN SIMPLE STRUCTURES.

CHAPTER IV. INTRODUCTORY.

§ i. Outline of Principles and Methods 53

PAGE

viii CONTENTS.

CHAPTER V. ROOF TRUSSES. — FRAMED STRUCTURES SUSTAINING
STATIONARY LOADS.

§ i. Loads on Roof Trusses 58

2. Roof Truss with Vertical Loads 61

3. Stresses Due to Wind Pressure 67

4. Maximum Stresses 71

5. Cases Apparently Indeterminate . 74

6. Three-hinged Arch . 80

7. Counterbracing 88

CHAPTER VI. SIMPLE BEAMS.

§ i. General Principles . 94

2. Beam Sustaining Fixed Loads 98

3. Beam Sustaining Moving Loads 101

CHAPTER VII. TRUSSES SUSTAINING MOVING LOADS.

§ i . Bridge Loads 112

2. Truss Regarded as a Beam 115

3. Truss Sustaining Any Series of Moving Loads 120

4. Truss with Subordinate Bracing 137

5. Uniformly Distributed Moving Load 145

PART III. — CENTROIDS AND MOMENTS OF INERTIA.

CHAPTER VIII. CENTROIDS.

§ i. Centroid of Parallel Forces 152

2. Center of Gravity — Definitions and General Principles . . . . 157

3. Centroids of Lines and of Areas 160

CHAPTER IX. MOMENTS OF INERTIA.

§ i. Moments of Inertia of Forces 167

2. Moments of Inertia of Plane Areas 177

CHAPTER X. CURVES OF INERTIA.

§ i. General Principles 187

2. Inertia-Ellipses for Systems of Forces 190

3. Inertia-Curves for Plane Areas 195

GRAPHIC STATICS,

PART I.
GENERAL THEORY.

CHAPTER L — DEFINITIONS. CONCURRENT
FORCES.

§ i. Preliminary Definitions.

1. Dynamics treats of the action of forces upon bodies. Its
two main branches are Statics and Kinetics.

Statics treats of the action of forces under such conditions
that no change of motion is produced in the bodies acted upon.

Kinetics treats of the laws governing the production of motion
by forces.

2. Graphic Statics has for its object the deduction of the
principles of statics, and the solution of its problems, by means
of geometrical figures.

3. A Force is that which tends to change the state of motion
of a body. We conceive of a force as a push or a pull applied
to a body at a definite point and in a definite direction. Such
a push or pull tends to give motion to the body, but this ten-
dency may be neutralized by the action of other forces. The
effect of a force is completely determined when three things
are given, — its magnitude, its direction, and its point of applica-
tion. The line parallel to the direction of the force and con-
taining its point of application, is called its line of action.

2 GRAPHIC STATICS.

Every force acting upon a body is exerted by some other body.
But the problems of statics usually concern only the body acted
upon. Hence, frequently, no reference is made to the bodies
exerting the forces.

4. Unit Force. — The ttnit force is a force of arbitrarily chosen
magnitude, in terms of which forces are expressed. Several
different units are in use. The one employed in this work is
the pound, which will now be defined.

A pound force is a force equal to the weight of a pound mass
at the earth's surface. A pound mass is the quantity of matter
contained in a certain piece of platinum, arbitrarily chosen, and
established as the standard by act of the British Parliament.

The pound force, as thus defined, is not perfectly definite,
since the weight of any given mass (that is, the attraction of
the earth upon it) is not the same for all positions on the earth's
.surface. The variations are, however, unimportant for most of
the requirements of the engineer.

In its fundamental meaning, the word "pound" refers to the
unit mass, and it is unfortunate that it is also applied to the unit
force. The usage is, however, so firmly established that it will
be here followed.

5. Concurrent and Non-concurrent Forces. — Forces acting on
the same body are concurrent when they have the same point of
application. When applied at different points they are non-
concurrent.

. 6. Complanar Forces are those whose lines of action are in
the same plane. In this work, only complanar systems will be
considered unless otherwise specified.

7. A Couple is the name given to a system consisting of two
forces, equal in magnitude, but opposite in direction, and having
different lines of action. The perpendicular distance between
the two lines of action is called the arm of the couple.

PRELIMINARY DEFINITIONS. 3

8. Equivalent Systems of Forces. — Two systems of forces
are equivalent when either may be substituted for the other
without change of effect.

9. Resultant. — A single force that is equivalent to a given
system of forces is called the resultant of that system. It will
be shown subsequently that a system of forces may not be
equivalent to any single force. When such is the case, the
simplest system equivalent to the given system may be called
its resultant. Any forces having a given force for their
resultant are called components of that force.

10. Composition and Resolution of Forces. — Having given
any system of forces, the process of finding an equivalent system
is called the composition of forces, if the system determined con-
tains fewer forces than the given system. If the reverse is the
case, the process is called the resolution of forces.

The process of finding the resultant of any given forces is
the most important case of composition ; while the process of
finding two or more forces, which together are equivalent to a
single given force, is the most common case of resolution.

11. Representation of Forces Graphically. — The magnitude
and direction of a force can both be represented by a line ; the
length of the line representing the magnitude of the force, and
its direction the direction of the force.

In order that the length of a line may represent the magni-
tude of a force, a certain length must be chosen to denote the
unit force. Then a force of any magnitude will be represented
by a length which contains the assumed length as many times
as the magnitude of the given force contains that of the unit
force.

In order that the direction of a force may be represented by
a line, there must be some means of distinguishing between the
two opposite directions along the line. The usual method is to
place an arrow-head on the line, pointing in the direction toward

4 GRAPHIC STATICS.

which the force acts. If the line is designated by letters placed
at its extremities, the order in which these are read may indicate
the direction of the force. Thus, AB and BA represent two
forces, equal in magnitude but opposite in direction.

The line of action of a force can also be represented by a line
drawn on the paper»

In solving problems in statics, it is usually convenient to
draw two separate figures, in one of which the forces are
represented in magnitude and direction only, and in the other
in line of action only.

These two species of diagrams will be called force diagrams
and space diagrams, respectively.

12. Notation. — The use of graphic methods is much facili-
tated by the adoption of a convenient system of notation in
the figures drawn.

There will generally be two figures (the force diagram and
the space diagram) so related that for every line in one there
is a corresponding line in the other.

In the force diagram each line represents a force in magni-
tude and direction ; in the space diagram the corresponding
line represents the line of action of the force. These lines
will usually be designated in the following manner : The line
denoting the magnitude and direction of
the force will be marked by two capital
letters, one at each extremity ; while the
action-line will be marked by the corre-
sponding small letters, one being placed at
each side of the line designated. Thus, in
Fig. i, AB represents a force in magni-
tude and direction, while its action-line is marked by the letters
abt placed as shown.

COMPOSITION OF CONCURRENT FORCES.

Fig. 3

§ 2. Composition of Concurrent Forces.

13. Resultant of Two Concurrent Forces. — If two concur-
rent forces are represented in magnitude and direction by two
lines AB and BC, their

resultant is represented
in magnitude and direc-
tion by AC. (Fig. 2.)
Proofs of this proposi-
tion are given in all
elementary treatises on
mechanics, and the demonstration will be here omitted. The
point of application of the resultant is the same as that of the
given forces. Thus if O (Fig. 2) is the given point of applica-
tion, then abt be, and ac, drawn parallel to AB, BC, and AC
respectively, are the lines of action of the two given forces and
their resultant. The figure marked (A) is a force diagram, and
(B) is the corresponding space diagram (Art. n).

14. Resultant of Any Number of Concurrent Forces. — If any
number of concurrent forces are represented in magnitude and
direction by lines AB, BC, CD, . . ., their resultant is repre-
sented in magnitude and direction by the line AN, where N is
the extremity of the line representing the last of the given
forces.

This proposition follows immediately from the preceding
one ; for the resultant of the forces represented by AB and
BC is a force repre- B

sented by AC\ the re-
sultant of AC and CD
is AD, and so on. By
continuing the process
we shall arrive at the
result stated. It is

readily seen that the order in which the forces are taken does
not affect the magnitude or direction of the resultant as thus

Fig. 3

6 GRAPHIC STATICS.

determined. The point of application of the resultant is the
same as that of the given forces. Figure 3 shows the force
diagram and space diagram for a system of four forces repre-
sented by ABt BCy CD, DE, and their resultant represented
by AE, applied at the point O.

From the above construction it is evident that every system
of concurrent forces has for its resultant some single force
(Art. 9) ; though in particular cases its magnitude may be zero.

15. Force Polygon. — The figure formed by drawing in suc-
cession lines representing in magnitude and direction any

number of forces is called a force polygon for
C n

those forces. Thus Fig. 4 is a force polygon

for any four forces represented in magnitude
and direction by the lines AB, BC, CD, and
DE, whatever their lines of action. It may
happen that the point E coincides with A, in
which case the polygon is said to be closed. It is evident that
the order in which the forces are taken does not affect the
relative positions of the initial and final points.

§ 3. Equilibrium of Concurrent Forces.

1 6. Definition. — A system of forces acting on a body is
in equilibrium if the motion of the body is unchanged by its
action.

17. Condition of Equilibrium. — In order that no motion
may result from the action of any system of concurrent forces,
the magnitude of the resultant must be zero ; and conversely,
if the magnitude of the resultant is zero, no motion can
result. But (Arts. 14 and 15) the condition that the result-
ant is zero is identical with the condition that the force
polygon closes. Hence, the following proposition :

If any system of concurrent forces is in equilibrium, the force
polygon for the system must close. And conversely, If the force

EQUILIBRIUM OF CONCURRENT FORCES. 7

polygon is closed for any system of concurrent forces, the system
is in equilibrium.

The comparison of this with the analytical conditions of
equilibrium is given in Art. 22.

1 8. Method of Solving Problems in Equilibrium. — If a sys-
tem of concurrent forces in equilibrium be partially unknown,
we may in certain cases determine the unknown elements by
applying the principles of Art. 17.

A common .case is that in which two forces are unknown
except as to lines of action. Thus, suppose a system of five
forces in equilibrium,
three being fully
known, represented in
magnitude and direc-
tion by AB, BC, CD
(Fig. 5), and in lines
of action by ab, be, cd,
while concerning the other two we know only their lines of
action de, ea. To determine these two in magnitude and
direction, it is necessary only to complete the force polygon of
which ABCD is the known part. The remaining sides must be
parallel respectively to de and ea. From D draw a line parallel
to de, and from A a line parallel to ea, prolonging them till they
intersect at E. Then DE and EA represent the required forces
in magnitude and direction, and the complete force polygon is
ABCDEA. It is evident that ABODE' A is an equally legiti-
mate form of the force polygon, and gives the same result for
the magnitude and direction of each of the unknown forces.
This problem occurs constantly in the construction of stress
diagrams by the method described in Part II.

The student will find little difficulty in treating other prob-
lems in the equilibrium of concurrent forces.

19. Problems in Equilibrium. — (i) A particle is in equilib-
rium under the action of five forces, three of which are

8 GRAPHIC STATICS.

completely known, while of the remaining two, one is known
in direction only, and the other in magnitude only. To deter-
mine the unknown forces.

(2) Suppose two forces known in magnitude but not in
direction, the remaining forces being wholly known.

(3) Suppose one force wholly unknown, the others being
known.

§ 4. Resolution of Concurrent Forces.

20. To Resolve a Given Force into Any Number of Com-
ponents having the same point of application, we have only to
draw a closed polygon of which one side shall represent the
magnitude and direction of the given force ; then the remaining
sides will represent, in magnitude and direction, the required
components. This problem is, in general, indeterminate, unless
the components are required to satisfy certain specified con-
ditions.

[NOTE. — A problem is said to be indeterminate if its conditions can be satisfied
in an infinite number of ways. It is determinate if it admits of only one solution.
Thus, the problem, to determine the values of x and y which shall satisfy the equation
x -\- y — 10, is indeterminate; while the problem, given 2^+3=7, to find the
value of x, is determinate. The case in which a finite number of solutions is possible
may be called incompletely determinate. Thus, the problem, given x2 + x — 6 — o,
to find x, admits of two solutions, and therefore is incompletely determinate. All
these classes of problems may be met with in statics.]

21. To Resolve a Given Force into Two Components. — This
problem is indeterminate unless additional data are given. For
if the given force be represented in magnitude and direction by
a line, any two lines which with the given line form a triangle
may represent forces which are together equivalent to the given
force. But an infinite number of such triangles may be drawn.
The solution of the following four cases of this problem will
form exercises for the student. In each case the force diagram
and space diagram should be completely drawn, and the student
should notice whether the problem is determinate, partially
determinate, or indeterminate.

RESOLUTION OF CONCURRENT FORCES.

(1) Let the lines of action of the required components be
given.

(2) Let the two components be given in magnitude only.

(3) Let the line of action of one component and the magni-
tude of the other be given.

(4) Let the magnitude and direction of one component be
given.

It will be noticed that these four cases correspond to four
cases of the solution of a plane triangle.

22. Resolved Part of a Force. — If a force is conceived to be
replaced by two components at right angles to each other, each
is called the resolved part ',* in its direction, of the given force.

It is readily seen that the resolved part of a force repre-
sented by AB (Fig. 6) in the direction of any line XX, is
represented in magnitude and direction
by A'B' , the orthographic projection of
AB upon XX. It follows that the
resolved part (in any given direction) of
the resultant of any concurrent forces
is equal to the algebraic sum of the
resolved parts of its components in that direction ; signs plus
and minus being given to the resolved parts to distinguish the
two opposite directions which they
may have. Thus (Fig. 7) the re-
solved parts of the forces AB, BC,
CD, in a direction parallel to XX,
.are A'B', B' C , CD'] and their
algebraic sum is A' D' , which is the
resolved part of the resultant AD.
If the resultant is zero, D1 coincides with Af ; hence,

(i) For the equilibrium of any concurrent forces, the sum
of their resolved parts in any direction must be zero.

Fig. 6

* The term " resolute " has been proposed by J. B. Lock (" Elementary Statics ") to
denote what is here defined as the resolved part of a force.

I0 GRAPHIC STATICS.

Again, if Dr coincides with Af, then either D coincides
with Ay or else AD is perpendicular to XX "; hence,

(2) If the sum of the resolved parts of any concurrent
forces in a given direction is zero, their resultant (if any)
is perpendicular to that direction. And if the sum of the
resolved parts is zero for each of two directions, the resultant
is zero, and the system is in equilibrium.

Propositions (i) and (2) state the conditions of equilibrium
for concurrent forces usually deduced in treatises employing
algebraic methods.

CHAPTER II. — NON-CONCURRENT FORCES.

§ i. Composition of Non-concurrent Forces Acting on the Same

Rigid Body.

23. Definition of Rigid Body. — A rigid body is one whose
particles do not change their positions relative to each other
under any applied forces. No known body is perfectly rigid,
but for the purposes of statics, most solid bodies may be con-
sidered as such ; and any body which has assumed a form of
equilibrium under applied forces, may, for the purposes of
statics, be treated as a rigid body without error.

24. Change of Point of Application. — The effect of a force
upon a rigid body will be the same, at whatever point in its
line of action it is applied, if only the particle upon which it
acts is rigidly connected with the body.

This proposition is fundamental to the development of the
principles of statics, and is amply justified by experience.*
In applying the principle, we are at liberty to assume a
point of application outside the actual body, the latter being
ideally extended to any desired limits.

25. Resultant of Two Non-Parallel Forces. — If two corn-
planar forces are not parallel, their lines of action must inter-
sect, and the point of intersection may be taken as the point
of application of each force. Hence, they may be treated as

* This proposition may be proved analytically by deducing the equations of motion
of a rigid body, and showing that the effect of any force on the motion of the body
depends only upon its magnitude, direction, and line of action. But such a proof is, of
course, outside the scope of this work.

12 GRAPHIC STATICS.

concurrent forces, and their resultant may be determined as
in Art. 13. The following proposition may therefore be
stated :

If two forces acting in the same plane on a rigid body
are represented in magnitude and direction by AB and BC,
their resultant is represented in magnitude and direction by

AC, and its line of action passes through the point of inter-
section of the lines of action of the given forces. Its point
of application may be any point of this line.

It may happen that the point of intersection of the two
given lines of action falls outside the limits available for the
drawing. In such a case it will be most convenient to find
the resultant by the method to be explained in Art. 27. The
same remark applies to the case of two parallel forces.

26. Resultant of Any Number of Non-concurrent Forces —
First Method. — The method of the preceding article may be
extended to the determination of the resultant of any number
of forces acting on the same rigid body. Let AB, BC, CD, DE

(Fig. 8), represent in
magnitude and direction
four forces, and let ab,
be, cd, de represent their
lines of action. To find
their resultant, we may
. . proceed as follows :

s % tf\ The resultant of AB

and BC is represented
in magnitude and direction by AC, and in line of action by ac
drawn parallel to AC through the point of intersection of ab
and be. Combining this resultant with CD, we get as their
resultant a force represented in magnitude and direction by

AD, and in line of action by ad drawn parallel to AD through
the point of intersection of ac and cd. This is evidently the
resultant of AB, BC, and CD. In the same way, this resultant

COMPOSITION OF NON-CONCURRENT FORCES. 13

combined with DE gives for their resultant a force whose mag-
nitude and direction are represented by AE, and whose line of
action is ae, parallel to AE and passing through the point in
which ad intersects de. This last force is the resultant of the
four given forces.

The process may evidently be extended to the case of any
number of forces.

As in the case discussed in the preceding article, this method
will become inapplicable or inconvenient if any of the points of
intersection fall outside the limits available for the drawing.
For this reason it is usually most convenient to employ the
method described in Art. 27.

The student should bear in mind that the length and direc-
tion AE and the line ae are not the magnitude, direction, and
line of action of any actual force applied to the body. By the
resultant is meant an ideal force, which, if it acted, would
produce the same effect upon the motion of the body as is
produced by the given forces. It is a force which may be
conceived to replace the actual forces, and may be assumed to
be applied to any particle in its line of action, provided that
particle is regarded as rigidly connected with the given body.
The line of action may in reality fail to meet the given body.
(See Art. 24.)

27. Resultant of Non-concurrent Forces — Second Method. —

This method will be described by reference to an example.
Referring to Fig. 9, let AB, BC, CD, DE represent in magni-
tude and direction four forces whose lines of action are ab, be,
cd, de ; and let it be required to find their resultant. Draw the
force polygon ABCDE, and from any point O in the force
diagram draw lines OA, OB, OC, OD, OE. These lines may
represent, in magnitude and direction, components into which
the given forces may be resolved. Thus AB is equivalent to
forces represented by AO and OB acting in any lines parallel
to AO, OB, whose point of intersection falls upon ab \ BC is

!4 GRAPHIC STATICS.

equivalent to forces represented by BO, OC, acting in any lines
parallel to BO, OC, which intersect on be ; and so for each of
the given forces. The four given forces may, therefore, be
replaced by eight forces given in magnitude and direction by

AO, OB, BO, OC, CO, OD, DO, OE, with proper lines of
action. Now, it is possible to make the lines of action of the
forces represented by OB and BO coincide ; and the same is
true of each pair of equal and opposite forces, OC, CO ; OD,
DO. To accomplish this, let AO, OB act in lines ao, ob, inter-
secting at any assumed point of ab. Prolong ob to intersect
be, and take the point thus determined as the point of inter-
section of the lines of action of BO, OC ; these lines are then
bo, oc. Similarly prolong oc to intersect cd, and let the point
of intersection be taken as the point at which CD is resolved
into CO and OD ; the lines of action of these forces are then
co and od. In like manner choose do, oe, intersecting on de, as
the lines of action of DO, OE. If this is done, the forces OB,
BO will neutralize each other and may be omitted from the
system ; also the pairs OC, CO, and OD, DO. Hence, there
remain only the two forces represented in magnitude and
direction by AO, OE, and in lines of action by ao, oe.
Their resultant is given in magnitude and direction by AE,
and its line of action is ae, drawn parallel to AE through the

COMPOSITION OF NON-CONCURRENT FORCES. 15

point of intersection of oa and oe ; and this is also the result-
ant of the given system.

By carefully following through this construction the student
will be able to reduce it to a mechanical method, which can
be readily applied to any system.

28. Funicular Polygon. — The polygon whose sides are oa, oby
oc, od, oe, is called & funicular polygon* for the given forces.

Since the point at which the two components of AB are
assumed to act may be taken anywhere on the line ab, there
may be any number of funicular polygons with sides parallel
to oa, ob, etc. Again, if O is taken at a different point, there
may be drawn a new funicular polygon starting at any point
of ab ; and by changing the starting point any number of
funicular polygons may be drawn with sides parallel to the
new directions of OA, OB, etc. Moreover, different force
and funicular polygons may be obtained by changing the order
in which the forces are taken.

It may be proved geometrically that for every possible funic-
ular polygon drawn for the same system of forces, the last
vertex, determined by the above method, will lie on the same
line parallel to the closing side of the force polygon (as ae,
Fig. 9). Such a proof is outside the scope of this work.
The truth of the proposition may, however, be shown from
the principles of statics. For if it were not true, it would
be possible by the above method to find two or more forces,
having different lines of action, which are equivalent to each
other, because each is equivalent to the given system. But
this is impossible.

29. Examples. — i. Choose five forces, assigning the magni-
tude, direction, and line of action of each, and find their
resultant by constructing the force and funicular polygons.

2. Draw a second funicular polygon, using the same point
O in the force diagram.

*Also called equilibrium polygon.

!6 GRAPHIC STATICS.

3. Draw a third funicular polygon, choosing a new point O.

4. Solve the same problem, taking the forces in a different
order.

30. Definitions. — The point O (Fig. 9) is called the pole of
the force polygon. The lines drawn from the pole to the
vertices of the force polygon may be called rays. The sides
of the funicular polygon are sometimes called strings.

Each ray in the force diagram is parallel to a corresponding
string in the space diagram. As a mechanical rule, it should
be remembered that the two rays drawn to the extremities of the
line representing any force are respectively parallel to the two
strings which intersect on the line of action of that force.

The rays terminating at the extremities of any side of the
force polygon represent in magnitude and direction two com-
ponents that may replace the force represented by that side ;
while the corresponding strings represent the lines of action of
these components. Thus (Fig. 9) the force represented by BC
may be replaced by two forces represented by BO, OC, acting
in the lines bo, oc.

The pole distance of any force is the perpendicular distance
from the pole to the line representing that force in the force
diagram. It may evidently be considered as representing the
resolved part, perpendicular to the given force, of either of the
components represented by the corresponding rays. Thus, OJ\f
(Fig. 9) is the pole distance of AB ; and OM represents the
resolved part, perpendicular to AB, of either OA or OB. The
student should notice particularly that the pole distance repre-
sents a force magnitude and not a length.

31. Forces not Possessing a Single Resultant. — It may hap-
pen that the first and last sides of the funicular polygon are
parallel, so that the above construction fails to give the line of
action of the resultant. This will be the case if the pole is
chosen on the line AE (Fig. 9), because the first and last sides
of the funicular polygon are respectively parallel to OA and

COMPOSITION OF NON-CONCURRENT FORCES. j^

OE. The difficulty will, in this case, be avoided by taking the
pole at some point not on the line AE. But in one particular
case OA and OE will be parallel, wherever the pole be taken.
By inspection of the force diagram it is seen that this can
occur only when the point E coincides with A. In this case
AO and OE represent equal and opposite forces; and unless
their lines of action coincide, they cannot be combined into a
simpler system. If their lines of action are coincident, the
forces neutralize each other and the resultant is zero ; if not,
the system reduces to a couple (Art. 7). If the lines of action
of. AO and OE coincide, they may still be regarded as forming
a couple, its arm being zero. Therefore,

If the force polygon for any system of forces closes, the result-
ant is a couple.

Now, it is evident that by shifting the starting point for the
funicular polygon, the lines of action of AO and OE will be
shifted ; and by taking a new pole, their magnitude, or direc-
tion,, or both, may be changed. Hence, there may be found
any number of couples, each equivalent to the same system
of forces, and therefore equivalent to each other. The relation:
between equivalent couples is discussed in Art. 52.

Example. — Assume five forces whose force polygon closes,
the lines of action being taken at random. Draw two funicular
polygons, using the same pole, and a third, using a different
pole ; thus reducing the given system to an equivalent couple
in three different ways.

32. Resultant Force and Resultant Couple. — From the above
discussion (Arts. 27 and 31), it is evident that any system of
complanar forces, acting on the same rigid body, is equivalent
either to a single force or to a couple. In other words, every
system of complanar forces possesses either a resultant force or a
resultant coziple. (See Art. 9.)

33. Comparison of Methods. — Of the methods given in Arts.
26 and 27, for finding the resultant of a system of non-concur-

18 GRAPHIC STATICS.

rent forces, the first is a special case of the second. For if the
pole in Fig. 9 be chosen at the point A, and the first string be
made to pass through the point of intersection of ab and be, the
construction becomes identical with that of Fig. 8. If the first
method is employed, we are liable to meet the difficulty men-
tioned in Arts. 25 and 26, that some of the required points of
intersection do not fall within convenient limits.

In the second method, since the pole may be chosen at pleas-
ure, the rays may generally be caused to make convenient angles
with the given forces, and all required points of intersection to
fall within convenient limits.

34. Closing of the Funicular Polygon. — Let the force poly-
gon be drawn for any given forces, and let A and E be the
initial and final points. Suppose a funicular polygon drawn,
corresponding to any pole O. The given system is equivalent
to two forces represented in magnitude and direction by AO,
OE, their lines of action being the first and last sides of the
funicular polygon. In general, these lines of action will not be
parallel; but, as explained in Art. 31, it may happen that they
are parallel. And if parallel, it may happen that they coincide.
In this case, the funicular polygon is said to be closed.

§ 2. Equilibrium of Non-concurrent Forces.

35. Conditions of Equilibrium. — From the foregoing princi-
ples the conditions of equilibrium for any system of complanar
forces may be deduced.

Whatever the system of forces may be, let a force polygon be
drawn, and let A and E be its initial and final points. Choose
a pole O, and draw a funicular polygon. The system is thus
reduced to two forces represented in magnitude and direction
by AO and OE\ their lines of action being the first and last
sides of the funicular polygon. In order that there may be
equilibrium, these two forces must be equal and opposite and
have the same line of action. In order that they may be equal

EQUILIBRIUM OF NON-CONCURRENT FORCES. 19

and opposite, A and E must coincide. In order that they may
have the same line of action, the first and last sides of the funic-
ular polygon must coincide. That is, in order that there may
be equilibrium, two conditions must be satisfied :

(a) TJie force polygon must close.

(#) The funicular polygon must close.

Conversely, if the force polygon closes and one funicular poly-
gon also closes, the system is in equilibrium. For, if the force
polygon closes, the two forces AO and OE are equal and oppo-
site; and if a funicular polygon closes, these two forces have
the same line of action and therefore balance each other.

It follows that if the force polygon and one funicular polygon
are closed, all funicular polygons will close.

36. Auxiliary Conditions of Equilibrium. — If any system of
forces in equilibrium be divided into two groups, the resultants
of the two groups are equal and opposite and have the same
line of action.

Particular case. — If three forces are in equilibrium, their
lines of action must meet in a point, or be parallel. For, if
the lines of action of two of the forces intersect, their result-
ant must act in a line passing through the point of inter-
section. But this resultant must be equal and opposite to
the third force and have the same line of action.

37. General Method of Solving Problems in Equilibrium. -

The principles of Art. 35 furnish a general method of solv-
ing, graphically, problems relating to the equilibrium of corn-
planar forces acting on a rigid body. The problems to be
solved will always be of the following kind :

A body is in equilibrium under the action of forces, some
of which are completely known, and others wholly or partly
unknown. It is required to determine the unknown elements.
The required elements may be either the magnitudes or the
lines of action of the forces.

GRAPHIC STATICS.

The general method of procedure is as follows : First, draw
the force polygon and funicular polygon so far as possible
from the given data ; then complete them, subject to the
condition that both polygons must close.

This general method will be illustrated by the solution of
several problems of frequent occurrence. We may here meet
with both determinate and indeterminate problems (Art. 20).
In order that a problem may be determinate, the given data,
together with the condition that the force polygon and funic-
ular polygon are to close, must be just sufficient to fully
determine these figures. There are many possible cases fur-
nishing determinate problems. Some of these cannot readily
be solved graphically. In the following articles are treated
three important cases to which the general method above
outlined is well adapted.

38. Problems in Equilibrium. I. — A rigid body is in equi-
librium under the action of a system of parallel forces, all
known except two, these being unknown in magnitude and
direction, but having known lines of action. It is required
to fully determine the unknown forces.

Let the known lines of action of five forces in equilibrium

be ab, be, cd, de, ea (Fig. 10), and let
AB, BC, CD represent the known
forces in magnitude and direction,
while DE and EA are at first un-
known. The force polygon, so far
as it can be drawn from the given
data, is the straight line ABCD.
Choose a pole O and draw rays OAy
OB, OC, OD. Parallel to these draw
in the space diagram the strings oay
ob, oc, od, four successive sides of the
funicular polygon. This much can
be drawn from the data given. We must now complete the

EQUILIBRIUM OF NON-CONCURRENT FORCES.

two polygons and cause both to close. In the funicular poly-
gon, but one side remains to be drawn; and in the force
polygon but one vertex remains unknown. If the forces are
taken in the order AB, BC, CD, DE, EA, the unknown vertex
in the force polygon is the one to be marked E ; the unknown
side of the funicular polygon is oe, and is to be parallel to
the line OE. But the string oe must pass through the inter-
section of od with de, and also through the intersection of
oa with ae. Hence, since these intersections are both known,
oe can be drawn at once as shown in the figure ; and then
OE can be drawn parallel to oe. This fixes the point E ;
.and DE and EA represent, in magnitude and direction, the
required forces whose lines of action are de, ea.

39. Problems in Equilibrium. II. — A rigid body is in
equilibrium under the action of a system of non-parallel forces,
all known, except two ; of these, the line of action of one
and the point of application of the other are given. It is
required to completely determine the unknown forces.

Let the system consist of five forces to be represented in
the usual way ; in magnitude and direction by lines marked
AB, BC, CD, DE, EA ; and in lines of action by ab, be, cd,
de, ea. Of these lines let be, cd, de, ea (Fig. 11), be given, and
let L be a given point of ab. Also let BC, CD, and DE be
known, while EA, AB are unknown. First, draw the force
polygon as far as possible, giving B, C, D, E, four consecutive
vertices of the force
polygon. The side EA
must be parallel to ea,
but its length is un-
known, hence the ver-
tex A of the force
polygon cannot be
fixed. Next, draw the
funicular polygon so

Fig. 11

22 GRAPHIC STATICS.

far as possible from the given data. Choose the pole O, and
draw the rays OB, OC, OD, GE. The remaining ray, OA,
cannot yet be drawn. Now, in the funicular polygon, the
sides ob, oc must intersect on be; oc and od must intersect
on cd\ od and oe must intersect on de ; oe and oa must inter-
sect on ea ; and oa and ob must intersect on ab. Since L
is the only known point of the line ab, let this point be
taken as the point of intersection of oa and ob. Now draw
ob through L parallel to OB, and successively oc, od, oe, par-
allel to OC, OD, OE. We may now draw oa, joining L with
the point in which oe intersects ea. This completes the funic-
ular polygon. The force polygon can now be completed as
follows : From O draw a line parallel to oa and from E a line
in the known direction of EA ; their intersection determines
A. This determines both EA and AB, and the force polygon
is completely known. The line of action ab may now be
drawn through L parallel to AB, and the forces are completely
determined.

40. Problems in Equilibrium. III. — A rigid body is in
equilibrium under the action of any number of forces, of
which three are known only in line of action ; the remaining
forces being completely known. It is required to determine
the unknown forces.

Let the given forces be six in number, their lines of action
being represented by ab, be, cd, de, ef, fa (Fig. 12), and let AB,
BC, and CD be known, while the remaining three forces are un-
known in magnitude and direction. The known data make it
possible to draw at once three sides of the force polygon,
namely, AB, BC, CD \ and the four rays OA, OB, OC, OD,
from any assumed pole O. Also, four sides of the funicular
polygon, oa, ob, oc, od, may be drawn at once. But oe and of
are unknown ; also DE, EF, and FA in the force polygon.

Since any side of the funicular polygon can be drawn through
any chosen point, let the polygon be started by drawing oa

EQUILIBRIUM OF NON-CONCURRENT FORCES. 23

through the intersection of efand fa. We may then draw suc-
cessively ob, oc, od. Now, of is unknown in direction ; but it
is to be drawn through the intersection of oa and fa, hence,
whatever its direction, it intersects ef in the same point (since
oa was drawn through the intersection of ef and fa}. Hence,
the two vertices of the funicular polygon falling on ef and fa
coincide at the intersection of these two lines. We may there-

fore draw oe through this point and also through the point
already found by the intersection of od and de. Now draw a
line from O parallel to oe, and from D a line parallel to de ;
their intersection is E. Draw from E a line parallel to ef, and
from A a line parallel to fa ; their intersection determines F.
The force polygon is now completely drawn, and DE, EF, FA
represent, in magnitude and direction, the required forces. The
remaining ray OF and the corresponding string of may now be
drawn, but are not needed.

The construction might have been made equally well by
choosing the intersection of de and ef as the starting point,
since two vertices of the funicular polygon may be made to
coincide at that point. Or, the point of intersection of de and
of might be chosen ; but in that case, the order of the forces
should be so changed as to make de and af consecutive. If
this were done the figures should be relettered to agree with

GRAPHIC STATICS.

the order in which the forces were taken. It may be noticed
that the direction of the string first drawn may be chosen arbi-
trarily and the pole so taken as to correspond to the direction
chosen. This is important in the treatment of the following
special case.

Case of inaccessible points of intersection. — It may happen
that the lines of action de, ef, and fa have no point of intersec-
tion within convenient limits. When this is the case, the
method just given may still be applied, but involves the geo-
metrical problem of drawing a line through an inaccessible point.
For example, if ef and fa intersect beyond the limits of the
drawing, as shown in Fig. 13, we may proceed as follows:
Choose some point of ab and from it draw a line through the
point of intersection of ef and fa. (This can be done by a
method to be explained presently.) Let this line be oa. From
the known point A of the force polygon draw a line parallel to

Fig. 13

ca and choose a pole upon it. Draw the rays OB, OC, OD ;
then the corresponding strings in the order ob, oc, od. From
the point in which od intersects de draw a line to the inaccessible
point of intersection of ef and fa ; this will be oe. The force
polygon can now be completed just as in the preceding case.

A line may be drawn through the inaccessible point of inter-
section of two given lines by the following method : Let AC,

EQUILIBRIUM OF NON-CONCURRENT FORCES.

BD (Fig. 14), be the given lines, and let it be required to draw

a line through their point of intersection from some point P.

Draw PA intersecting AC and

BD, and from C, any point of

AC, draw a line CO, parallel to

PA. From E, a point of BD,

draw EA, EP ; also draw CF

parallel to AE, and FQ parallel

to EP. Then PQ is the line

required. For, by the similar

triangles, it is easily shown that FIS. 14

— = ; which proves that AC, BG, and PQ meet in a point.

Exception. — If the lines of action of the three unknown
forces meet in a point (or are parallel), the problem is impos-
sible of solution, unless it happens that the resultant of the
known forces acts in a line through the point of intersection
of those three lines of action (or parallel to them) ; in which
case the problem is indeterminate. These cases will not be
discussed here.

Remark. — In the problems treated in this and the two pre-
ceding articles, it will be noticed that the forces should be taken
in such an order that those which are completely known are
consecutive. Otherwise the number of unknown lines in the
force and funicular polygons will be increased. [For an-
other method of solving this problem, see Levy's " Statique
Graphique."]

41. Examples. — I. A rigid beam AB rests horizontally
upon supports at A and B, and sustains loads as follows :
Its own weight of 100 Ibs. acting at its middle point ; a load
of 50 Ibs. at C ; a load of 60 Ibs. at D ; and a load of 80 Ibs.
at E. The successive distances between A, C, D, E, and B
are 4 ft., 6 ft., 7 ft., and 10 ft. Required the upward pressures
on the beam at the supports.

26 GRAPHIC STATICS.

2. A rigid beam AB is hinged at A, and rests horizontally
with the end B upon a smooth horizontal surface. The beam
sustains loads as in Example I, and an additional force of 40
Ibs. is applied at the middle point at an angle of 45° with the
bar in an upward direction. Required the pressures upon the
beam at A and B. [The pressure at A may have any direc-
tion, while the pressure at B must be vertically upward, i.e.,
at right angles to the supporting surface. Hence, this is a
particular case of Problem II.]

3. Let the end B of the bar rest against a smooth surface
making an angle of 30° with the horizontal ; the remaining
data being as in Example 2.

4. A rigid bar 2 ft. long weighs 10 Ibs., its center of gravity
being 8 inches from one end. The bar rests inside a smooth
hemispherical bowl of 15 inches radius. What weight must
be applied at the middle point in order that the bar may rest
when making an angle of 15° with the horizontal? Also,
what are the reactions at the ends ?

5. A uniform bar 20 inches long, weighing 10 Ibs., rests
with one end against a smooth vertical wall and the other
end overhanging a smooth peg 10 inches from the wall. A
weight P is suspended from the end so that the bar is in
equilibrium when making an angle of 30° with the horizontal.
Find P, and the pressures exerted on the bar by the wall
and peg.

42. Special Methods. — Certain problems can be treated
more simply by other methods than by the general method
of constructing the force and funicular polygons. This is
sometimes true of the following :

Problem. — A rigid body is held in equilibrium by four
forces acting in known lines, only one being known in magni-
tude and direction. It is required to completely determine
the three remaining forces. (See Fig. 15.)

Let the four forces have lines of action ab, be, cd, da, and let

EQUILIBRIUM OF NON-CONCURRENT FORCES. 2/

AB be drawn representing the magnitude and direction of the
known force. Now the resultant of the forces whose lines
of action are da and ab must act in a line passing through
M, the point of intersection of these lines ; and the resultant
of the other two forces must act in a line passing through
Nt the point of intersection of be and cd. But the two

resultants must be equal and opposite and have the same
line of action, else there could not be equilibrium. (Art. 36.)
Hence, each must act in the line MN. Draw BD parallel
to MN, and AD parallel to ad\ the point D being determined
by their intersection. Then DA represents in magnitude and
direction the force acting in da, and DB the resultant of
DA and AB. But BD must represent the resultant of the
two remaining forces ; hence these two forces are represented
by BC and CD drawn from B and D parallel respectively to
be and cd.

This problem is a special case of that treated in Art. 40.
But the construction here given will in many cases be the
simpler one.

Example. — A rigid body has the form of a square ABCD,
the side AB being horizontal, and BC vertical. The weight
of the body is 100 Ibs., its center of gravity being at the inter-
section of the diagonals. It is held in equilibrium by three
forces as follows : P^ acting at C in the line AC; P2 acting at
D in the line AD ; and P3 applied at B and acting in the line
joining B with the middle point of AD. Required to com-
pletely determine Ply P2, and P3.

2g GRAPHIC STATICS.

§ 3. Resolution into Non-concurrent Systems.

43. To Replace a Force by Two Non-concurrent Forces. —

This may be done in an infinite number of ways. The lines
of action of the two components must intersect at a point on
the line of action of the given force, and they must further
satisfy the same conditions as concurrent forces. (Art. 21.)

44. To Replace a Force by More Than Two Non-concurrent
Forces. — This may be done by first resolving the given force
into two forces by the preceding article, and then resolving
one or both of its components in the same way. This problem
and that of Art. 43 are indeterminate. (See note, Art. 20.)
To make such a problem determinate, something must be
specified regarding the magnitudes and lines of action of the
required components. We shall consider some of the particular
cases which are of frequent use in the treatment of practical
problems.

45. Resolution of a Force into Two Parallel Components. —

Let it be required to resolve a force into two components
having given lines of action parallel to its own.

If the given force be reversed in direction, it will form with
the required components a system in equilibrium. The com-
ponents may then be determined by the method of Art. 3-8.

Example. — Let the student solve two particular cases of
this problem, taking the line of action of the given force (i)
between those of the components, (2) outside those of the
components.

46. Resolution of a Force into Three Components. — Problem.
— To resolve a force into three components having known lines
of action.

If the given force be reversed in direction, it will form, with
the required forces, a system in equilibrium. Hence these
forces may be determined by either of the two methods given

RESOLUTION INTO NON-CONCURRENT SYSTEMS.

in Arts. 40 and 42. Or, the following reasoning may be
employed, leading to the same construction as that of Art. 42.

Let AD (Fig. 16) represent the given force in magnitude
and direction, and ad its line of action ; the lines of action of
the components being given as ab, be, cd. Since the given
force may be assumed to act at any point in the line ad, let its
point of application be taken at M, the point of intersection of

Fig. 16

ad and cd. Resolve it into two components acting in the lines
cd and MN, N being the point of intersection of ab and be.
These components are represented in magnitude and direction
by AC, CD. Let AC, acting in the line MN (also marked ac\
be resolved into components having lines of action ab, be.
These components are given in magnitude and direction by
AB, BC, drawn parallel respectively to ab, be. Hence, the
given force, acting in ad, is equivalent to the three forces
represented in magnitude and direction by AB, BC, CD, acting
in the lines ab, be, cd.

If the line of action of the given force does not intersect any
one of the given lines ab, be, cd, within the limits of the draw-
ing, it may be replaced by two components ; then each may be
resolved in accordance with the above method, and the results
combined. If the three lines ab, be, cd, are all parallel to ad,
or if the four lines intersect in a point, the problem is inde-
terminate. This is evident from the preceding articles, since,
by methods already given, a part of AD can be replaced by
two components acting in any two of the given lines, as ab,
be ; and the remaining part by two components acting in ab, cd;

30 GRAPHIC STATICS.

or in be, cd. This construction evidently admits of infinite
variation.

For another method of solving the above problem, see Clarke's
"Graphic Statics," p. 16.

§ 4. Moments of Forces and of Couples.

47. Moment of a Force. — Definition. — The moment of a
force with respect to a point is the product of the magnitude of
the force into the perpendicular distance of its line of action
from the given point. The moment of a force with respect to
an axis perpendicular to the force is the product of the magni-
tude of the force by the perpendicular distance from the axis
to the line of action of the force.

If the moment is taken with respect to a point, that point is
called the origin of moments. The perpendicular distance from
the origin, or axis, to the line of action of the force is called
the arm.

Rotation tendency of a force. — The moment of a force
measures its tendency to produce rotation about the origin, or
axis. Thus, if a rigid body is fixed at a point, but free to turn
about that point in a given plane, any force acting upon it in
that plane will tend to cause it to rotate about the fixed point.
The amount of this tendency will be proportional both to the
magnitude of the force and to the distance of its line of action
from the given point ; that is, to the moment of the force with
respect to the point, as above defined.

Rotation in any plane may have either of two opposite
directions, which may be distinguished from each other by signs
plus and minus. Rotation with the hands of a watch supposed
placed face upward in the plane of the paper will be called
negative, and the opposite kind positive. It would be equally
legitimate to adopt the opposite convention, but the method
here adopted agrees with the usage of the majority of writers.

The sign of the moment of a force is regarded as the same
as that of the rotation it tends to produce about the origin.

MOMENTS OF FORCES AND OF COUPLES. 31

Moment represented by the area of a triangle. — If a triangle
be constructed having for its vertex the origin of moments, and
for its base a length in the line of action of a force, numerically
equal to its magnitude, then the moment of the force is numeri-
cally equal to double the area of the triangle. This follows at
once from the definition of moment.

48. Moment of Resultant of Two Non-parallel Forces. — Propo-
sition. — The moment of the resultant of two non-parallel forces
with reference to a point in their plane is equal to the algebraic
sum of their separate moments with reference to the same point.

In Fig. 17, let AB, BC, and AC represent in magnitude and
direction two forces and their resultant ; and let ab, be, ac be
their lines of action, intersecting in a
point N. Choose any point M as the
origin of moments. Lay off NP = AB\
NQ = BC\ and NR = AC. Then the
moments of the three forces are re-
spectively equal to double the areas of
the triangles MNP, MNQ, MNR.
These three triangles have a common
side MN, which may be considered the
base ; hence their areas are propor-
tional to their altitudes measured from that side. These alti-
tudes are PP', QQ' , RR', perpendicular to MN. Now, if PP"
is parallel to MN, P"R' is equal to PP' ; also RP" equals
QQ' (since they are homologous sides of the equal triangles
NQQ, PRP"). Hence RR' = PP' + QQ', and therefore

Area J/A^-area J/./VP -f- area MNQ ;

which proves the proposition.

If the origin of moments is so taken that the moments of
AB and BC have opposite signs, the demonstration needs modi-
iication. The student should attempt the proof of this case for
himself.

32 GRAPHIC STATICS.

49. Moment of a Couple. — Definition. — The moment of a
couple about any point in its plane taken as an origin is the
algebraic sum of the moments of the two forces composing it
with reference to the same origin.

Proposition. — The moment of a couple is the same for every
origin in its plane, and is numerically equal to the product of
the magnitude of either force into the arm of the couple.
(Art. 7.)

The proof of this proposition can readily be supplied by the
student.

50. Moment of the Resultant of Any System. — Proposition.
-The algebraic sum of the moments of any given complanar

forces, with reference to any origin in their plane, is equal to
the moment of their resultant force or resultant couple with
reference to the same origin.

The construction employed in proving this proposition is
similar to that used in Art. 27, which the student may profit-
ably review at this point. Referring to Fig. 9, let the given
forces be represented in magnitude and direction by AB, BC,
CD, DE, and in lines of action by ab, be, cd, de. Let the
origin of moments be any point in the space diagram. As in
Art. 27, replace AB by AO, OB, acting in lines ao, ob ; replace
BC by BO, OC, acting in lines bo, oc ; replace CD by CO, OD,
acting in lines co, od; and replace DE by DO, OE, acting in
lines do, oe. (For brevity, we refer to a force as AB, meaning
"the force represented in magnitude and direction by AB.1')

Now by Art. 48, we have, whatever the origin,

Moment of AB = moment of AO-}- moment of OB;

" BC= " " BO+ " " OC;

« CD= " " CO + " " OD;

" DE= " " DO+ " " OE.

Since the forces represented by BO and OB have the same
line of action, their moments are numerically equal but have

MOMENTS OF FORCES AND OF COUPLES.

opposite signs ; and similar statements are true of CO and
OC, and of DO and OD. Hence, the addition of the above
four equations shows that the sum of the moments of AB,
BC, CD, DE, is equal to the sum of the moments of AO
and OE. Now the given system has either a resultant force
or a resultant couple. In the first case the resultant of the
system is the resultant of AO and OE, and its moment is
equal to the algebraic sum of their moments, by Art. 48.
In the second case (which occurs only when E coincides with
A), the resultant couple is composed of AO and OE (which
in this case are equal and opposite forces), and the moment
of the couple is, by definition, equal to the algebraic sum of
the moments of AO and OE. Hence, in either case, the
proposition is true.

It should be noticed that the proof here given applies to
systems of parallel forces, as well as to non-parallel systems.
The proposition of Art. 48 could be extended to the case of
any number of forces, by considering first the resultant of
two forces, then combining this resultant with the third force,
and so on ; but the method would fail if, in the process, it
became necessary to combine parallel forces. The method
here adopted is not subject to this failure.

51. Condition of Equilibrium. — It follows from what has
preceded, that if a given system is in equilibrium, the alge-
braic sum of the moments 'must be zero, whatever the origin.
For, in case of equilibrium, AO and OE (Fig. 9) are equal and
opposite and have the same line of action ; hence, the sum
of their moments (which is the same as the sum of the
moments of the given forces) is equal to zero. Conversely,.
If the algebraic sum of the moments is zero for every origin,
the system must be in equilibrium. For, if it is not, there
is either a resultant force or a resultant couple. But the
moment of a force is not zero for any origin not on its line
of action ; and the moment of a couple is not zero for any

GRAPHIC STATICS.

origin. For a fuller discussion of the conditions of equilibrium,
see Arts. 58 and 59.

52. Equivalent Couples. — Proposition. — If a system has for
its resultant a couple, it is equivalent to any couple whose
moment is equal to the sum of the moments of the forces
of the system.

For, as already seen (Art. 31), when the resultant is a
couple, the force polygon is closed. Let the initial and final
points of the force polygon coincide at some point A, and
let O be the pole. Then the forces of the resultant couple
are represented in magnitude and direction by AO, OA. Since
the position of O is arbitrary, the force AO (or OA) may be
made anything whatever in magnitude and direction. Also
the line of action of the force AO may be taken so as to pass
through any chosen point. Hence, the resultant couple may
have for one of its forces any force whatever in the plane of
the given system ; and the other force will have such a line
of action that the moment of the couple will be equal to the
sum of the moments of the given forces.

This reasoning is equally true if the given system is a
couple. Hence, a couple is equivalent to any other couple
having the same moment. In other words, all couples whose
•moments are eqtial are equivalent ; and conversely, all equiva-
lent couples have equal moments.

[NOTE. — The construction above discussed fails if all forces of the given system
are parallel to the direction chosen for the forces of the resultant couple. For then
the force polygon is a straight line, and if the pole is chosen in that line, the
strings of the funicular polygon are parallel to the lines of action of the forces, and
the polygon cannot be drawn. But in this case, the system may first be reduced to a
couple whose forces have some other direction, and this couple may be reduced to
one whose forces have the direction first chosen. Hence, the proposition stated
holds in all cases.]

53. Moment of a System. — Definition. — The moment of a
system of forces is the algebraic sum of the moments of the
forces of the system.

GRAPHIC DETERMINATION OF MOMENTS. 35

54. Moments of Equivalent Systems. — Proposition. — The
moments of any two equivalent systems of complanar forces
with respect to the same origin are equal.

This follows immediately from the preceding articles. For,
if the two systems are equivalent, each is equivalent to the
same resultant force or resultant couple, and the moments of
the two systems are therefore each equal to the moment of
this resultant and hence to each other.

§ 5 . Graphic Determination of Moments.

55. Proposition. — If, through any point in the space dia-
gram a line be drawn parallel to a given force, the distance
intercepted upon it by the two strings corresponding to that
force, multiplied by the pole distance of the force, is equal
to the moment of the force with respect to the given point.

By the strings " corresponding to " a given force are meant
the two strings which intersect at a point on its line of action.

Let AB (Fig. 18) represent the magnitude and direction of
a force whose line of action is ab, and let M be the origin
of moments. Let O be the pole,
and OK ( = H) the pole distance
of the given force. Draw the
strings 0a, ob, and through M
draw a line parallel to aby inter-
secting oa and ob in P and Q.
Then it is to be proved that the

moment of the given force with respect to M is equal to
HxPQ.

Let h equal the perpendicular distance of M from ab. Then
the required moment is AB x h. But since the similar triangles
OAB and RPQ have bases AB, PQ, and altitudes H, //, respec-
tively, it follows that

^ = ~: hence,
which proves the proposition.

36 GRAPHIC STATICS.

It should be noticed that PQ represents a length, while H
represents a force magnitude. Hence, the moment of the given
force with respect to M is equal to the moment of a force H
with an arm PQ. [It may, in fact, be shown that the given
force is equivalent to an equal force acting in the line PQ
(whose moment about M is therefore zero), and a couple with
forces of magnitude H, and arm PQ. For AB acting in ab is
equivalent to AO and OB acting in ao and ob respectively.
Also, AO acting in ao is equivalent to forces represented by
AK and KO acting respectively in PQ and in a line through
P at right angles to PQ ; and OB may be replaced by forces
represented by OK and KB, the former acting in a line through
Q perpendicular to PQ, and the latter in PQ. But AK and
KB are equivalent to AB ; hence, the proposition is proved.]

56. Moment of the Resultant of Several Forces. — The mo-
ment of the resultant of any number of consecutive forces in
the force and funicular polygons may be found by a method
similar to that just described. Thus, let Fig. 19 represent the

force polygon and the
funicular polygon for six
forces, and let it be re-
quired to find the moment
of the resultant of the four
forces represented in the
force polygon by BC, CD,
DE, and EF, with respect
to any point M. The re-
sultant of the four forces
is represented by BF, and

acts in a line through the intersection of ob and of. Through
M draw a line parallel to BF, intersecting ob and of in P and Q
respectively. Then PQ multiplied by OK, the pole distance of
BF, gives the required moment. This method does not apply
to the determination of the moment of the resultant of several
forces not consecutive in the force polygon.

SUMMARY OF CONDITIONS OF EQUILIBRIUM. 37

57. Moments of Parallel Forces. — The method of Arts. 55
,and 56 is especially useful when it is desired to find the mo-
ments of any or all of a system of parallel forces ; since with
such a system the pole distance is the same for all forces, and
the moments are therefore proportional to the intercepts found
by the above method.

Example. — Assume five parallel forces at random ; choose
an origin, and determine their separate moments, also the
moment of their resultant, by the method of Arts. 55 and 56.

§ 6. Summary of Conditions of Equilibrium.

58. Graphical and Analytical Conditions of Equilibrium Com-
pared.— It has been shown (Art. 35) that the conditions of
equilibrium for a system of complanar forces acting on a rigid
body are two in number :

(I) The force polygon must close.

(II) Any funicular polygon must close.

The analytical conditions * are the following :

(1) The algebraic sum of the resolved parts of the forces
in any direction must be zero.

(2) The algebraic sum of their moments for any origin must
be zero.

The condition (i) is readily seen to be equivalent to (I).
For if the sides of the force polygon be orthographically
projected upon any line, their projections will represent in
magnitude and direction the resolved parts of the several
forces parallel to the line ; and, if the force polygon is closed,
the algebraic sum of these projections is zero, whatever the
direction of the assumed line. (See Art. 22.) It may also
be seen that condition (II) carries with it (2). For, if every
funicular polygon closes, the system is equivalent to two equal
and opposite forces having the same line of action (Arts. 31,

* See Minchin's Statics, Vol. I, p. 114.

38 GRAPHIC STATICS.

35, and 50) ; and the sum of the moments of these two forces
must be zero.

A further comparison may be made. The analytical condi-
tion (2) carries (i) with it ; and similarly the graphical condi-
tion (II) carries with it the condition (I). That (2) includes
(i) may be seen as follows : If the sum of the moments is zero
for one origin Mit there can be no resultant couple, neither can
there be a resultant force unless with a line of action passing
through Mv If the sum of the moments is zero for a
second origin M%, the resultant force, if one exists, must
act in the line M1Af2. If the sum of the moments is zero
also for a third origin MS, not on the line M\M* there can
be no resultant force. It follows at once that condition (i)
must hold.

That (II) includes (I) may be shown as follows : Let A and
E be the first and last points of the force polygon, and choose
a pole O\. Then, if the funicular polygon closes, dA and
OiE are parallel. Choose a second pole O2, and, if the funic-
ular polygon again closes, O2A and OZE are parallel. AE
must then be parallel to OiOz, unless A and E coincide.
Now, choose a third pole O3, not on O\O^\ if the funicular
polygon for this pole closes, O3A and O3E must be parallel.
But this is impossible unless A and E coincide ; that is, unless
condition (I) holds.

The last result may be reached in another way. With any
pole O draw a funicular polygon and suppose it to close. The
system is thus reduced to two forces acting in the same line
oa. Hence, there is no resultant couple, and if there is a
resultant force, its line of action is oa. With the same pole
draw a second funicular polygon, the first side being o'a\
parallel to oa. If this polygon closes, there can be no result-
ant force, for if one existed it would act in the line o'a' ; and
there would thus be two resultant forces, acting in different
lines oa and o'a\ which is impossible.

SUMMARY OF CONDITIONS OF EQUILIBRIUM. 39

59. Summary. — It is now evident that the conditions neces-
sary to insure equilibrium may be stated in several different
ways, both analytically and graphically. To summarize :

A. Analytically : There will be equilibrium if either of the
following conditions is satisfied :

(1) The sum of the moments is zero for each of three points
not in the same line.

(2) The sum of the moments is zero for each of two points,
and the sum of the resolved parts is zero in a direction not
perpendicular to the line joining those two points.

(3) The sum of the moments is zero for one point, and the
sum of the resolved parts is zero for each of two directions.

B. Graphically : There will be equilibrium if either of these
three conditions is satisfied :

(1) A funicular polygon closes for each of three poles not in
the same line.

(2) Two funicular polygons close for the same pole.

(3) One funicular polygon closes and the force polygon closes.

CHAPTER III. INTERNAL FORCES AND
STRESSES.

§ i . External and Internal Forces.

60. Definitions. — It was stated in Art. 3 that every force
acting upon any body is exerted by some other body. In what
precedes, we have been concerned only with the effects pro-
duced by forces upon the bodies to which they are applied.
It has therefore not been needful to consider the bodies which
exert the forces. It is now necessary to consider forces in
another aspect.

The forces applied to any particle of a body may be either
external or internal.

An external force is one exerted upon the body in question
by some other body.

An internal force is one exerted upon one portion of the body
by another portion of the same body.

It is important to note, however, that the same force may be
internal from one point of view, and external from another.
Thus, if a given body be conceived as made up of two parts,
X and Y, a force exerted upon X by Y is internal as regards
the whole body, but external as regards the part X. Thus, let

AB (Fig. 20) represent
a bar, acted upon by two
forces of equal magnitude

. SO

applied at the ends paral-
lel to the length of the bar, in such a way as to tend to pull it
apart. These two forces are exerted upon AB by some other

EXTERNAL AND INTERNAL FORCES. 4I

bodies not specified. If the whole bar be considered, the
external forces acting upon it are simply the two forces
named.

But suppose the body under consideration is AC, a portion of
AB. The external for.ces acting upon this body are (i) a force
at A, already mentioned, and (2) a force at C, exerted upon AC
by CB. This latter force is internal to the bar AB, but external
to AC.

61. Conditions of Equilibrium Apply to External Forces. — In

applying the conditions of equilibrium deduced in previous
articles, it must be remembered that only external forces are
referred to. It is also important to notice that the principles
apply to any body or any portion of a body in equilibrium ; and
the system of forces in every case must include all forces that
are external to the body or portion of a body in quqstion.

Thus, if the bar AB (Fig. 20) be in equilibrium under the
action of two opposite forces P and Q, applied at A and B
respectively, as shown in the figure, the principles of equilibrium
may be applied either to the whole bar, or to any part of it, as
AC.

(a) For the equilibrium of the whole bar AB, we must have
P=Q, these being supposed the only external forces acting on
the bar.

(b) For the equilibrium of AC, the force exerted upon AC by
CB t must be equal and opposite to P. This latter force is
external^ AC, though internal^ the whole bar.

The method just illustrated is of frequent use in the investi-
gation of engineering structures. It is often desired to deter-
mine the internal forces acting in the members of a structure,
and the general method is this : Direct the attention to such a
portion of the whole structure or body considered that the
internal forces which it is desired to determine shall be external
to the portion in question. (See Art. 67.)

42 GRAPHIC STATICS.

§ 2. External and Internal Stresses.

62. Newton's Third Law. — Let X and Y be any two portions
of matter ; then if X acts upon Y with a certain force, Y acts
upon X with a force of equal magnitude in the opposite direc-
tion. This is the principle stated in Newton's third law of
motion, — that " to every action there is an equal and contrary
reaction." It is justified by universal experience.

63. Stress. — Definition. — Two forces exerted by two por-
tions of matter upon each other in such a way as to constitute
an action and its reaction, make up a stress.

Illustrations. — The earth attracts the moon with a certain
force, and the moon attracts the earth with an equal and opposite
force. The two forces constitute a stress.

Two electrified bodies attract (or repel) each other with equal
and opposite forces. These two forces constitute a stress.

Any two bodies in contact exert upon each other equal and
opposite pressures (forces), constituting a stress.

By the magnitude of a stress is meant the magnitude of either
of its forces.

64. External and Internal Stresses. — It has been seen that
two portions of matter are concerned in every stress. Now the
two portions may be regarded either as separate bodies, or as
parts of a body or system of bodies which includes both.

A stress acting between two parts of the same body (or
system of bodies) is an internal stress as regards that body
or system.

A stress acting between two distinct bodies is an external
stress as regards either body.

It is important to notice that the same stress may be internal
from one point of view, and external from another. Thus, if
a given body be considered as made up of two parts X and F,
a stress exerted between X and F is internal to the whole body,
but external to either X or F.

EXTERNAL AND INTERNAL STRESSES.

Illustration. — Consider a body AB (Fig. 21) resting upon a
second body Y, and supporting another body X, as shown. If
the weight of the body AB be disregarded,
the forces acting upon it are (i) the down-
ward pressure (say P) exerted by X at the
surface A, and the upward pressure (say Q)
exerted by Fat B ; these forces being equal
and opposite, since the body is in equilibrium.
Now the body X is acted upon by AB with
a force equal and opposite to P, and these
two forces constitute a stress which is external to AB. There
is also an external stress exerted between AB and Y at B.
But let AB be considered as made up of two parts, AC and CB.
Then (Art. 60) CB exerts upon AC a force upward, and AC
exerts upon CB a force downward. These two forces are an
action and its reaction, and constitute a stress which is internal
to the body AB. This same stress is, however, external to
either AC or CB. An equivalent stress evidently exists at
every section between A and B. (When we refer to the force
acting upon CB at C, we mean the resultant of all forces exerted
upon the particles of CB by the particles of AC. This resultant
is made up of very many forces acting between the particles.
Also the stress at C means the stress made up of the two
resultants of the forces exerted by AC and CB upon each other.)

65. Three Kinds of Internal Stress. — It is evident that the
internal stress at C in the body AB (Fig. 20) depends upon
the external forces applied to the body. If the forces at A and
B cease to act, the forces exerted by AC and CB upon each
other become zero. If the forces at A and B are reversed in
direction, so also are those at C. (As a matter of fact, the
particles of AC exert forces upon those of CB, even if the
external forces do not act. But if the external forces applied
to CB are balanced, the resultant of the forces exerted on CB
by A C is zero.)

44 GRAPHIC STATICS.

The nature of the internal stress at any point in a body is
thus seen to depend upon the external forces applied to the body.

Now, if we consider two adjacent portions of a body (as
the parts X and Y, Fig. 22) separated by a plane surface, the

external forces may have either of three
tendencies : (i) to pull X and Y apart
in a direction perpendicular to the plane
of separation ; (2) to push them together

in a similar direction ; (3) to slide each over the other along the
plane of separation. Corresponding to these three tendencies,
the stress between X and Y may be of either of three kinds :
tensile, compressive, or shearing.

A tensile stress is such as comes into action to resist a
tendency of the two portions of the body to be pulled apart in
the direction of the normal to their surface of separation.

A compressive stress is such as comes into action to resist a
tendency of X and Y to move toward each other along the
normal to the surface.

A shearing stress is such as acts to resist a tendency of X
and Fto slide over each other along the surface between them.

In case of a tensile stress, the force exerted by X upon Y
has the direction from Y toward X\ and the force exerted by
Y upon X has the direction from X toward Y.

In case of a compressive stress, the force exerted by X upon
Fhas the direction from X toward F; and the force exerted by
Fupon X has the direction from F toward X.

In the case of a shearing stress, the force exerted by X
upon F may have any direction in the plane of separation ; the
force exerted by Fupon X having the opposite direction.

If X and F are separate bodies, instead of parts of one
body, a similar classification may be made of the kinds of
stress between them ; but with these we shall have no occasion
to deal. The terms tensile stress, compressive stress, and sJiear-
ing stress (or tension, compression, and sJiear) are usually applied
only to internal stresses.

EXTERNAL AND INTERNAL STRESSES.

66. Strain. — In what has preceded, the bodies dealt with
have been regarded as rigid ; that is, the relative positions of
the particles of any body have been regarded as remaining
unchanged. But, as remarked heretofore, no known body is
perfectly rigid. If no external forces act upon a body, its
particles take certain positions relative to each other, and the
body has what is called its natural shape and size. If external
forces are applied, the shape and size will generally be changed ;
the body is then said to be in a state of strain. The deforma-
tion produced by any system of applied forces is called the
strain due to those forces. The nature of this strain in any
case depends upon the way in which the forces are applied.- It
is unnecessary to treat this subject further at this point, since
we shall at present be concerned only with problems in the
treatment of which it will be sufficiently correct to regard the
bodies as rigid.

[NOTE. — There is a lack of uniformity among writers in regard to the meanings
attached to the words stress and strain. It may, therefore, be well to explain again
at this point the way in which these words are used in the following pages. The
word stress should be employed only in the sense above defined, as consisting of two
equal and opposite forces constituting an action and its reaction. The two forces are
exerted respectively by two bodies or portions of matter upon each other. An
internal stress is a stress between two parts of the same body. An internal force is
one of the forces of an internal stress. It is intended in what follows to use the
word? " internal stress " (or simply " stress ") only when both the constituent forces
are referred to; and when only one of the forces is meant, to use the words "inter-
nal force" (or simply "force"). It will be noticed, therefore, that in the following
pages the words "force in a bar " are frequently used where many writers would say
'• stress." This departure from the usage of many high authorities seems justified by
the following considerations : (i) It agrees with the usage which is being adopted
by the highest authorities in pure mechanics. (2) It is desirable that the nomencla-
ture of technical mechanics shall agree with that of pure mechanics, so far as they
deal with the same conceptions. The definition of strain above given is in conform-
ity with the usage of the majority of the more recent text-books. But it is not rare
to find in technical literature the word strain used in the sense of internal stress as
above defined. Such use of the word should be avoided.]

46 GRAPHIC STATICS.

§ 3. Determination of Internal Stresses.

67. General Method. — The stresses exerted between the
parts of a body may or may not be completely determinate by
means of the principles already deduced. But in all cases
these principles suffice for their partial determination. The
general method employed is always the same, and will now be
illustrated. As heretofore we deal only with complanar forces.
Let XY (Fig. 23) represent a body in equilibrium under the
action of any known external forces as shown. Now conceive
the body to be divided into two parts as X and Y, separated
by 'any surface. The particles of X near the surface exert
upon those of Y certain forces, and are in return acted

upon by forces exerted by
the particles of Y. These
forces are internal as re-

^ gards the whole body. In

P6 order to determine them so

• *eC»

far as possible, we proceed

as follows : Let the resultant of all the forces exerted by Y
upon JTbe called T; then T is either a single force or a couple.
Now apply the conditions of equilibrium to the body X. The
external forces acting on X are P^ P2, P3, and 7! Since Plt P2,
and Ps are supposed known, T can be determined. In fact,
T is equal and opposite to the resultant of Plt P2, and P3.

So much can always be determined. But T is the resultant
of a great number of forces acting on the various particles
of X\ and these separate forces cannot in general be deter-
mined by methods which lie within the scope of this work.

The general principle just illustrated may be stated as
follows :

If a body in equilibrium under any external forces be conceived
as made up of two parts X and Y, then the internal forces
exerted by X upon Y, together with the external forces acting on
K, form a system in equilibrium.

DETERMINATION OF INTERNAL STRESSES. 47

As an immediate consequence, we may state that the result-
ant of the forces exerted by X upon Y is equivalent to the result-
ant of the external forces acting on X ; and is equal and opposite
to the resultant of the external forces acting ttpon Y.

Example. — Assume a bar of known dimensions, and the
magnitudes, directions, and points of application of five forces
acting on it. Then (i) determine a sixth force which will
produce equilibrium ; and (2) assume the bar divided into two
parts and find the resultant of the forces exerted by each part
on the other.

68. Jointed Frame. — In certain ideal cases (corresponding
more or less closely to actual cases), the internal forces may be
more completely determined. The most important of these
cases is that which will be now considered. Conceive a rigid
body made up of straight rigid bars hinged together at the ends.
Assume the following conditions :

(1) The hinges are without friction.

(2) All external forces acting on the body are applied at
points where the bars are joined together.

The meaning of these conditions will be seen by reference to
Fig. 24. The three bars X, Y, and Z are connected by a "pin
joint," the end of each bar having a hole
or "eye" into which is fitted a pin.
(Of course the three bars cannot be in
the same plane, but they may be nearly
so, and will be so assumed in what fol-
lows.) Condition (i) is satisfied if the
pin is assumed frictionless. The effect of this is that the force
exerted upon the pin by any bar (and the equal and opposite
reaction exerted upon the bar by the pin) act in the normal
to the surfaces of these bodies at the point of contact ; and,
therefore, through the centers of the pin and the hole. Condi-
tion (2) means that any external force (that is, external to the

GRAPHIC STATICS.

whole body) applied to any bar is applied to the end and in a
line through the center of the pin.

With the connection as shown, the bars do not exert forces
upon each other directly. But each exerts a force upon the
pin, and any force exerted by Y or Z upon the pin causes an
equal force to be exerted upon X. (This is seen by applying"
the condition of equilibrium to the pin.) Hence, in considering
the forces acting upon any bar as X, we may disregard the pin
and assume that each of the other bars acts directly upon X.
By what has been said, all such forces exerted upon X by the
other bars meeting it at the joint may be regarded as acting at
the same point — the center of the pin. We therefore treat the
bars as mere " material lines," and regard all forces exerted on
any bar (whether by the other bars or by outside bodies) as
applied at the ends of this " material line."

Since, with these assumptions, all forces acting on any bar
MN (Fig. 25) are applied either at M or N, the forces applied
at M must balance those applied at N. The resultants of the
two sets must therefore be equal and opposite, and have the
same line of action — namely MN. Further, it follows that
the stress in the bar, acting across any plane perpendicular to
its length, is a direct tension or compression.

69. Internal Stresses in a Jointed Frame. — Let Fig. 25 rep-
resent a jointed frame such as above described, in equilibrium

under any known external forces.
Let us apply the general method
of Art. 67 to this case. Divide
the body into two parts, X and Y9
by the surface AB as shown. Now
apply the conditions of equilibrium
to the body X. The system of
forces acting upon this body consists of PI, P2, P6, and the forces
exerted by Fupon X m the three members cut by the surface
AB. By Art. 68 the lines of action of these forces are known,

Fig- &i&

DETERMINATION OF INTERNAL STRESSES.

being coincident with the axes of the members cut. Hence,
the system in equilibrium consists of six forces, three com-
pletely known, and three known only in lines of action.
The determination of the unknown forces in magnitude and
direction is then a case under the general problem discussed in
Art. 40.

Nature of the stresses. — As soon as the direction of the
force acting upon X in any one of the members cut is known,
the nature of the stress in that member (whether tension or
compression) is known. For a force toward X denotes com-
pression ; while a force away from X denotes tension. (Art.

6S.)

If a section can be taken cutting only two members, the
forces in these may be found by the force polygon alone. The
same is true, if any number of members are cut, but the stresses
in all but two are known.

The methods described in the last three articles will find
frequent application in the chapters on roof and bridge trusses,
Part II.

Example. — Assume a jointed frame similar to the one shown
in Fig. 25, and let external forces act at all the joints. Then
(i) assume all but three of the forces known in magnitude and
direction and determine the remaining three so as to produce
equilibrium. (2) Take a section cutting three members and
determine the stresses in those members.

70. Indeterminate Cases. — If, in dividing the frame, more
than three members are cut, the number of unknown forces is
too great to admit of the determination of their magnitudes.
In such a case, it may happen that a section elsewhere through
the body will cut but three members ; and that after the deter-
mination of the stresses in these three, another section can be
taken cutting but three members whose stresses are unknown.
So long as this can be continued, the determination of the

50 GRAPHIC STATICS.

internal stresses can proceed. Thus, in Fig. 26, if a section
be first taken at AB, there are four unknown forces to be
determined. But, if the section A' B' be first taken, the
stresses in the three members cut may be determined ; after

which the section AB will
\A' introduce but three unknown

stresses.

There may, however, be
cases in which the stresses
cannot all be determined by
any method. With such ac-
tually indeterminate cases we shall not usually have to deal. It
should be noticed, also, that even when only three members
are cut, the problem is indeterminate if these tJiree intersect in
a point. As in the case just discussed, this indeterminateness
may be either actual or only apparent ; in the latter case it may
be treated as above indicated.

No attempt is here made to develop all methods that are
applicable or useful in the determination of stresses in jointed
frames. Some of these are best explained in connection with
the actual problems giving rise to them. We have sought here
-only to explain and clearly illustrate general principles.

71. Funicular Polygon Considered as Jointed Frame. — Let

.ab, be, cd, da (Fig. 27) be the lines of action of four forces in
equilibrium, the force polygon being ABCDA. Choosing a

pole, draw any funicular polygon, as the one shown. Now let
the body upon which the forces act be replaced by a jointed

DETERMINATION OF INTERNAL STRESSES. 51

frame whose bars coincide with the sides of the funicular
polygon. If at the joints of this frame the given forces be
applied, the frame will be in equilibrium ; and each bar will
sustain a tension or compression whose magnitude is repre-
sented by the corresponding ray of the force diagram.

To prove this, we apply the "general method" of Art. 67.
Consider any joint (as the intersection of oa and ob), and let
the frame be divided by a plane cutting these two members.
Then the portion of the frame about the joint is acted upon
by three forces : AB, acting in the line ab, and forces acting in
the bars cut, their lines of action being oa, ob. If the bar
oa sustains a compression and ob a tension, their magnitudes
being represented by OA and BO respectively, the portion
of the frame about the joint will be in equilibrium. Hence,
the tendency of the force AB is to produce the stresses men-
tioned in the bars oa, ob. In the same way it may be shown
that the tendency of the force BC is to produce in ob and
oc tensile stresses of magnitudes OB and CO, respectively.
Applying the same reasoning to each joint, it is seen that every
part of the frame will be in equilibrium if the bars sustain
stresses as follows : The bar oa must sustain a compression
OA ; ob a tension OB; oc a tension OC\ and od a. compression
OD. Hence, if the bars are able to sustain these stresses, the
frame will be in equilibrium.

If the stress in any member of the frame is a tension, that
member may be replaced by a flexible string. This is the
origin of the name string as applied to the sides of the funic-
ular polygon. This name is retained for convenience, but, as
just shown, it is not always appropriate.

72. Outline of Subject. — The foregoing pages, embracing
Part I, have been devoted to a development of the principles
of pure statics. We pass next to the application of these
principles to special classes of problems.

Part II treats of the determination of internal stresses in

52 GRAPHIC STATICS.

engineering structures. Only "simple" structures are consid-
ered,— that is, those whose discussion does not involve the
theory of elasticity. The structures considered include roof
trusses, beams, and bridge trusses.

Part III develops the graphic methods of determining cen-
troids (centers of gravity) and moments of inertia of plane
areas, including a short discussion of " inertia-curves."

PART II.
STRESSES IN SIMPLE STRUCTURES.

CHAPTER IV. INTRODUCTORY.

§ I. Outline of Principles and Methods.

73. The Problem of Design. — When any structure is sub-
jected to the actior. of external forces, there are brought into
action certain internal stresses in the several parts of the struc-
ture. The nature and magnitudes of these stresses depend upon
the external forces acting (Art. 65). In designing the structure,
each part must be so proportioned that the stresses induced in
it will not become such that the material cannot sustain them
without injury.

To determine these internal stresses, when the external forces
are wholly or partly given, is the problem of design, so far as
it will be here treated.

74. External Forces. — The external forces acting on a struc-
ture must generally be completely known before the internal
stresses can be determined. These external forces are usually
only partly given, and the first thing necessary is to determine
them fully.

The external forces include (i) the loads which the structure
is built to sustain, and (2) the reactions exerted by other bodies
upon the structure at the points where it is supported. The
former are known or assumed at the outset and the latter are
to be determined.

54 GRAPHIC STATICS.

75. Two Classes of Structures. — Structures may be divided
into two classes, according as they may or may not be treated
as rigid bodies in determining the reactions. This may be
illustrated as follows :

Let a bar AB (Fig. 28) be supported in a horizontal position
at two points A and B, the supports being smooth so that the

pressures on the bar at
A and B are vertical.
W&//M Let a known load P be

applied to the beam at a
given point, and let it be required to determine the reactions at
A and B.

This is a determinate problem ; for there are three parallel
forces in equilibrium, two being known only in line of action.
This problem was solved in Art. 38.

But let AC (Fig. 29) be a rigid bar supported at three points,,
A, B, and C, the reactions at those points being vertical.
IP |P2 |p Let any known loads

— ^ ±1 C be applied at given

points, and let it be
required to determine
Fig. so the three reactions.

This problem is indeterminate ; for any number of sets of
values of the three reactions may be found, which, with the
applied loads, would produce equilibrium if acting on a rigid
body. (See Art. 40.)

Since, however, an actual bar is not a perfectly rigid body,
such a problem as the one just stated is, in reality, determinate.
But it cannot be solved without making use of the elastic
properties of the material of which the body is composed.

The two classes of problems are, therefore, the following :
(i) those in which the reactions can be determined by treating
the structure as a rigid body, and (2) those in which the
determination of the reactions involves the theory of elasticity.
We shall at present deal only with the former class of prob-

OUTLINE OF PRINCIPLES AND METHODS. 55

lems. Structures coming under this class will be called simple
structures.

76. Truss. — A truss is a structure made up of straight bars
with ends joined together in such a manner that the whole acts
as a single body. The ends of the bars are, in practice, joined
in various ways ; but in determining the internal stresses, the
connections are assumed to be such that no resistance is offered
by a joint to the rotation of any member about it. Such a
structure to be indeformable must be made up of triangular
elements ; for more than three bars hinged together in the
form of a polygon cannot constitute a rigid whole. If the
external forces are applied to the truss only at the points where
the bars are joined, the internal stress at any section of a mem-
ber will be a simple tension or compression, directed parallel to
the length of the bar. (See Art. 68.)

The most important classes of trusses are roof trusses and
bridge trusses. The methods used in discussing these classes,
are, of course, applicable to any framed structures under similar
conditions.

77. Loads on a Truss. — The loads sustained by a truss may
be either fixed or moving. A fixed (or dead) load is one whose
point of application and direction remain constant. A moving
(or live) load is one whose point of application passes through
a series of positions. Fixed loads may be either permanent or
temporary.

The loads on a roof truss are usually all fixed, but are of
various kinds, viz., the weight of the truss itself and of the roof
covering, which is a permanent load ; the weight of snow
lodging on the roof, and the pressure of wind, both of which
are temporary loads.

A bridge truss supports both fixed and moving loads. The
former include the weight of the truss itself, of the roadway,
of all lateral and auxiliary bracing (permanent loads) ; and of
snow (a temporary load). The latter consist of moving trains.

56 GRAPHIC STATICS.

in the case of railway bridges, and of teams or crowds of animals
or people in the case of highway bridges.

78. Combination of Stresses Due to Different Causes. — When
a truss is subject to a variety of external loads, it is often
convenient to consider the effect of a part of them separately.
If tensile and compressive stresses are distinguished by signs
plus and minus, the stress in any member due to the combined
action of any number of loads is equal to the algebraic sum of
the stresses due to the loads acting separately.

A proof of this proposition might be given ; but it may be
accepted as sufficiently evident without formal demonstration.

79. Beams. — Another class of bodies to be treated is
included under the name beam.

A beam may be defined as a bar (usually straight) resting on
supports and carrying loads. The loads and reactions are
commonly applied in a direction transverse to the length of the
bar ; but this is not necessarily the case.

The internal stresses in any section of a beam are less
simple than those in the bars of an ideal jointed frame such as
a truss is assumed to be. A discussion of beams is given in
Chap. VI.

80. Summary of Principles Needed. — It will be well to
summarize at this point the main principles and methods which
will be employed in the discussion of the problems that follow.

The general problem presented by any structure consists
of two parts : (a) the determination of the unknown external
forces (or reactions) and (b) the determination of the internal
stresses.

(a) In the case of simple structures the unknown reactions
are usually two in number, and the cases most commonly pre-
sented are the following :

ist. — Their lines of action are known and parallel.

2nd. — The line of action of one and the point of application
of the other are known.

OUTLINE OF PRINCIPLES AND METHODS. 57

Since all the external forces form a system in equilibrium,
these two cases fall under the general problems discussed in
Arts. 38 and 39.

(b) In the case of a jointed frame or truss, the lines of
action of all internal forces are known, since they coincide
with the axes of the truss members. (Art. 68.) In determin-
ing their magnitudes we may have to deal with the following
problems in equilibrium :

ist. — The system in equilibrium may be completely known,
except the magnitudes of two forces.

2nd. — The magnitudes of three forces may be unknown.

The first case may be solved by simply making the force
polygon close. (See Art. 35.)

The second case may be solved by the method of Art. 40,
which consists in making the force and funicular polygons
close ; or by the method of Art. 42 ; or by the principle of
moments (Art. 51).

The student should be thoroughly familiar with the prob-
lems and principles here referred to. In the following chapters'
we proceed to their application.

8 1. Division of the Subject. — The subject of the design of
structures, so far as here dealt with, will be treated in three
divisions. The first relates to framed structures sustaining
only stationary loads ; the second to beams sustaining both
fixed and moving loads ; the third to framed structures sustain-
ing both fixed and moving loads. Among structures of the
first class, the most important are roof trusses ; hence, these
arc chiefly referred to in the next chapter. For a similar
reason, the chapter devoted to the third class of structures
refers principally to bridge trusses. The chapter on beams
precedes that on bridge trusses, for the reason that the methods
used in dealing with a beam under moving loads form a useful
introduction to those employed in treating certain classes of
truss problems.

CHAPTER V. ROOF TRUSSES. —FRAMED STRUC-
TURES SUSTAINING STATIONARY LOADS.

§ i. Loads on Roof Trusses.

82. Weights of Trusses. — Among the loads to be sustained
by a roof truss is the weight of the truss itself. Before the
structure is designed, its weight is unknown. But, since it is
necessary to know the weight in order that the design may be
correctly made, the method of procedure must be as follows :

Make a preliminary estimate of the weight, basing it upon
knowledge of similar structures ; or, in the absence of such
knowledge, upon the best judgment available. Then design
'the various truss members, compute their weight, and com-
pare the actual weight of the truss with the assumed weight.
If the difference is so great as to materially affect the design of
the truss members, a new estimate of weight must be made,
and the computations repeated or revised. No more than one
or two such trials will usually be needed.

As a guide in making the preliminary estimate of weight,
the following formulas may be used. They are taken from Mer-
riman's " Roofs and Bridges," being intended to represent
approximately the data for actual structures, as compiled by
Ricker in his " Construction of Trussed Roofs."

Let /=span in feet ; a = distance between adjacent trusses in
feet; W= total weight of one truss in pounds. Then for
wooden trusses

W-tal(i+lgi);

and for wrought iron trusses

W=$al(i+^l).

LOADS ON ROOF TRUSSES.

83. Weight of Roof Covering. — The weight of roof covering
can be correctly estimated beforehand from the known weights
of the materials. The following data may be employed, in the
absence of information as to the specific material to be used.
(See Merriman's " Roofs and Bridges," p. 4.) The numbers
denote the weight in pounds per square foot of roof surface.

Shingling : Tin, I Ib. ; wooden shingles, 2 to 3 Ibs. ; iron, I
to 3 Ibs. ; slate, 10 Ibs. ; tiles, 12 to 25 Ibs.

Sheathing : Boards I in. thick, 3 to 5 Ibs.

Rafters : 1.5 to 3 Ibs.

Purlins : Wood, I to 3 Ibs. ; iron, 2 to 4 Ibs.

Total roof covering, from 5 to 35 Ibs. per square foot of roof
surface.

84. Snow Loads. — The weight of snow that may have to be
borne will differ in different localities. For different sections
of the United States the following may be used as the maximum
snow loads likely to come upon roofs.

Maximum for northern United States, 30 Ibs. per square foot
of horizontal area covered.

For latitude of New York or Chicago, 20 Ibs. per square foot.

For central latitudes in the United States, lolbs. per square foot.

The above weights are given in Merriman's " Roofs and
Bridges." They are in excess of those used by some Bridge
and Roof companies.

85. Wind Pressure Loads. — The intensity of wind pressure
against any surface depends upon two elements : (a) the velocity
of the wind, and (b) the angle between the surface and the
direction of the wind.

Theory indicates that the intensity of wind pressure upon a
surface perpendicular to the direction of the wind should be
proportional to the square of the velocity of the wind relative to
the surface. As an approximate law this is borne out by
experiment. If / denotes the pressure per unit area, and v
the velocity of the wind, the law is expressed by the formula

GRAPHIC STATICS.

p — kv'2'. Here k is proportional to the density of air. Its
numerical value may be taken as 0.0024, if the units of force,
length, and time are the pound, foot, and second respectively.

If the wind strikes a surface obliquely, experiment shows
that the resulting pressure has a direction practically normal to
the surface. The tangential component is inappreciable, owing
to the very slight friction between air and any fairly smooth
surface. The intensity of the normal pressure depends upon
the angle at which the wind meets the surface.

For a given velocity of wind let /a denote the normal pressure
per unit area, when the direction of wind makes an angle a
with the surface, and /„ the pressure per unit area due to the
same wind striking a surface perpendicularly. Then the follow-
ing formula * has been given :

2 sin

It will rarely be necessary to use values of pn greater than
50 Ibs. per square foot. The following table gives values of
the coefficient of pn in the above formula for different values
of a. The value of pn may be taken as from 40 to 50 Ibs. per
square foot.

2 sin a

I + sin2 a

0°

0.00

50°

0.97

10°

o-34

60°

0.99

20°

0.61

70°

I.OO

30°

0.80

80°

I.OO

40°

0.91

90°

I.OO

* This formula is given by various writers. It is cited by Langley (" Experiments in
Aerodynamics," p. 24), who attributes it to Duchemin. Professor Langley's elaborate
experiments show so close an agreement with the formula that it may be used without hesi-
tation in estimating the pressure on roofs.

ROOF TRUSS WITH VERTICAL LOADS. 6l

§ 2. Roof Truss with Vertical Loads.

86. Notation. — The method of determining internal stresses
in the case of vertical loading will be explained by reference to
the form of truss shown in Fig. 30. The method will be seen
to be independent of the particular form of the truss.

For designating the truss members and the lines of action of
external forces a notation will be employed similar to that used
in previous chapters. Let each of the areas in the truss dia-
gram be marked with a letter or other symbol as shown in
Fig. 30 ; then the truss member or force-line separating any
two areas may be designated by the two symbols belonging to
those areas. Thus, the lines of action of the external forces
are ab, be, cd, etc., and the truss members are gk, hb, hi, etc.
It is to be noticed that the lines representing the truss mem-
bers represent also the lines of action of forces, — namely, the
internal forces in the members. The joint, or point at which
several members meet, may be designated by naming all the
surrounding letters. Thus, bcih, hijg are two such points.

87. Loads and Reactions. — The loads now considered are
assumed to be applied in a vertical direction, and to act at the
upper joints of the truss. This assumption as to the points of
application may in some cases represent very nearly the facts ;
in other cases the loads will, in reality, be applied partly at
intermediate points on the truss members. If the latter is the
case, the load borne upon any member is assumed to be divided
between the two joints at its ends. In this case the member
will be subject not only to direct tension or compression, but to
bending. With the latter we are not here concerned, although
it must always be considered in designing the member.

The ends of the truss are supposed to be supported on hori-
zontal surfaces, and the reaction at each point of support is
assumed to have a vertical direction.

If the loading is symmetrical with reference to a vertical line

62 GRAPHIC STATICS.

through the middle of the truss, it is evident that each reaction
is equal to half the total load. If the loading is not symmetri-
cal, the reactions cannot be determined so simply. They may,
however, be readily computed by either graphic or algebraic
methods. Graphically, the problem is identical with that solved
in Art. 38. The truss is treated as a rigid body, the external
forces acting upon it being the loads and reactions, which form
a system of parallel forces in equilibrium. Two of these forces
(the reactions) are unknown in magnitude, but known in line
of action. The construction for determining their magnitudes
is as follows :

Draw the force polygon ABCDEF for the five loads ; choose
a pole O, draw rays OA, OB, OC, etc., and draw the funicular
polygon as shown in Fig. 30. The two polygons are to be
completed by including the reactions FG, GA, and both poly-
gons must close. We may draw first og, the closing line of the
funicular polygon, and then the ray OG parallel to it, thus
determining the point G in the force polygon. The reactions
are now shown in magnitude and direction by FG and GA.

\
88. Determination of Internal Stresses. — When the external

forces are all known, the internal stresses may be found very
readily. The only principle needed is, that for any system of
forces in equilibrium, the force polygon must close. The con-
struction will now be explained.

Considering any joint of the truss (Fig. 30) as gkijg, fix the
attention upon the portion of the truss bounded by the broken
line in the figure. This portion is a body in equilibrium under
the action of four forces whose lines of action coincide with the
axes of the four bars gh, hi, if, jg, respectively. These forces
are internal as regards the truss as a whole, but external to the
part in question ; each force being one of the pair constituting
the internal stress at any point of the bar. Such a force acts
from the joint if the stress in the bar is a tension ; toward it if
the stress is a compression. (Art. 69.)

ROOF TRUSS WITH VERTICAL LOADS. 63

Since these four forces form a system in equilibrium, their
force polygon must close. This condition will enable us to fully
determine the magnitudes of the forces, provided all but two are
known, since the polygon can then be constructed as in Art. 18.

We cannot, however, begin with the joint just considered,
since at first the four forces are all unknown in magnitude.

If, however, we start with the joint gabh, the polygon of
forces can be at once drawn. For, reasoning as above, it is
seen that the portion of the truss immediately surrounding
this joint is in equilibrium under the action of four forces :
the reaction in the line ga, the load in the line ab, and the
internal forces in the lines bh, hg. Of these forces, two (the
reaction and the load) are completely known ; and it is neces-
sary only to draw a polygon of which two sides represent the
known forces and the other two sides are made parallel to the
members bht hg. Such a polygon is shown in Fig. 30 ; and
BH and HG represent in magnitude and direction the forces
whose lines of action are bh, hg.

64 GRAPHIC STATICS.

Evidently, BH and HG represent also the magnitudes (Art.
63) of the internal stresses in the two members bh and Jig.
The nature of these stresses may be found as follows : Since
the four forces represented in the polygon GABHG are in
equilibrium, and since GA, AB are the directions of two of
them, the directions of the other two must be BH, HG.
Hence, BH acts toward the joint and HG from it. This
shows that the stress in bh is a compression, while that in hg is
a tension.

Passing now to the joint bcihb, it is seen that of the four
forces whose lines of action meet there, two are fully known,
namely, the load BC acting vertically downward and the inter-
nal force in hb acting toward the joint (since the stress is com-
pressive), while the remaining two (viz., the internal forces in
ci and ih) are unknown in magnitude and direction. Since,
however, the unknown forces are but two in number, the force
polygon can be completely drawn, and is represented by the
quadrilateral HBCIH. The directions of the forces are found
as in the preceding case, and it is seen that the bars ci and ih
both sustain compressive stresses.

The process may be continued by passing to the remaining
joints in succession, in such order that at each there remain to
be determined not more than two forces. The complete con-
struction is shown in Fig. 30.

It is evident that the loads AB and EF might have been
omitted without changing the stresses in any of the truss
members. For their omission would leave as the complete
force polygon for external forces BCDEGB, and the two
reactions would be GB and EG ; but the force diagram would
be otherwise unchanged.

The great convenience of the notation adopted is now seen.*

*This is known as Bow's notation. The notation adopted in Part I involves the same
idea, but it is not usually employed in works on Graphic Statics, though possessing very
evident advantages. It was suggested to the writer by its use in certain of Professor
Eddy's works.

ROOF TRUSS WITH VERTICAL LOADS. 65,

The line representing the stress in any member is designated
in the force diagram by letters similar to those which designate
that member in the truss diagram. The latter is evidently a
space diagram (Art. n). The force diagram is often called a
stress diagram, since it shows the values of the internal stresses
in the truss members.

89. Reciprocal Figures. — There are always two ways of com-
pleting the force polygon when two of the forces are known
only in lines of action. (See Art. 18.) Either way will give
correct results, but unless a certain way be chosen, it will
become necessary to repeat certain lines in the stress diagram.
Thus, if, in Fig. 30, instead of GABHG we draw GABH'G,
the lines GH\ H1B are not in convenient positions for use in
the other polygons of which they ought to form sides. The
lettering of the diagrams will also be complicated. As an aid
in drawing the lines in the most advantageous positions, it is
convenient to remember the fundamental property of figures
related in such a way as the force and space diagrams shown in

Fig- 30.

Such figures are said to be reciprocal with regard to each
other. The fundamental property of reciprocal figures is that
for every set of lines intersecting in a point in either figure,,
there is in the other a set of lines respectively parallel to them
and forming a closed polygon.

It is also an aid to remember that the order of the sides in
any closed polygon in the stress diagram is the same as the
order of the corresponding lines in the truss diagram, if taken
consecutively around the joint. This usually enables us at
once to draw the sides of each force polygon in the proper
order.

90. Order of External Forces in Force Polygon. — It will be
observed that in the case above considered, in constructing the
force polygon for the loads and reactions, these forces have
been taken consecutively in the order in which their points of

GRAPHIC STATICS.

application occur in the perimeter of the truss. This is a
necessary precaution in order that the stress diagram and truss
diagram may be reciprocal figures, so that no line in the former
need be duplicated.

This requirement should be especially noticed in such a case
as that shown in Fig. 31, in which loads are applied at lower as

well as at upper joints. If the reactions are found by the
method of Art. 87, without modification, the force polygon will
not show the external forces in the proper order, since the
known forces are not applied at consecutive joints of the truss.
A new polygon should therefore be drawn after the reactions
have been determined.

If desirable (as in some cases it may be) to make use of a
funicular polygon in which the external forces are taken con-
secutively, this may be drawn after the reactions are deter-
mined and the new force polygon is drawn.

If a load be applied at some joint interior to the truss, as at
M (Fig. 31), then in constructing the stress diagram it should

STRESSES DUE TO WIND PRESSURE. 6/

be assumed to act at N, where its line of action intersects the
exterior member of the truss, and the fictitious member MN
inserted. The stresses in the actual truss members will be
unaffected by this assumption, and such a device is necessary in
order that the stress diagram may be the true reciprocal of the
truss diagram.

§ 3. Stresses Due to Wind Pressure.

91. Direction of Reactions Due to Wind Pressure. — Since
the effective pressure of the wind has the direction normal to
the surface of the roof (Art. 85), it has a horizontal component
which must be resisted by the reactions at the supports.
These cannot, therefore, act vertically, as in the case when
the loading is vertical. Their actual directions will depend
upon the manner in which the ends of the truss are sup-
ported.

If the ends of the truss are immovable, the directions of the
reactions cannot be determined, since any one of an infinite
number of pairs of forces acting at the ends would produce
equilibrium. (The same would be true of the reactions due to
vertical loads.) In such a case the usual assumption is one of
the following : (i) the reactions are assumed parallel to the
loads ; (2) the resolved parts of the reactions in the horizontal
direction are assumed equal.

In the case of trusses of large span it is not unusual to
support one end of the truss upon rollers so that it is free to
move horizontally, the other end being hinged, or otherwise
arranged to prevent both horizontal and vertical motion. This
allows for expansion and contraction with change of tempera-
ture, as well as for movements due to the small distortions of
the truss under loads. With this arrangement the reaction at
the end supported on rollers must be vertical ; and since the
point of application of the other reaction is known, both can
be fully determined by the method described in Art. 39.

68 GRAPHIC STATICS.

92. Determination of Reactions. — The methods of finding
reactions will now be explained for the three cases mentioned in
the preceding article : (i) Assuming both reactions parallel to the
wind pressure ; (2) assuming the horizontal resolved parts of the
two reactions equal ; and (3) assuming one reaction vertical.

(1) The first case needs no explanation, since it is identical
with that described in Art. 87, except that the loads and
reactions have a direction normal to one surface of the roof,
instead of being vertical. It is to be noticed that this assump-
tion cannot be made if the roof surface is curved, since the
lines of action of the forces will not be parallel. But since
the direction of the resultant of the loads will be known from
the force polygon, both reactions may be assumed to act par-
allel to this resultant, and the construction made as before.

(2) In the second case, let each reaction be replaced by two
forces acting at the support, one horizontal and the other
vertical. The two horizontal forces are known as soon as the
force polygon for the loads is drawn, and the two vertical
forces may be found as in the preceding case, since their lines
of action are known.

In Fig. 32, let ab, be, cd be the lines of action of the wind
forces. Let the right reaction be considered as made up of a
horizontal component acting in de and a vertical component
acting in ef \ and let the left reaction be replaced by a vertical
component acting in fg and a horizontal component acting in
ga. Draw the force polygon (or "load-line") ABCD. By the
assumption already made GA and DE are to be equal, and
their sum is to equal the horizontal resolved part of AD.
Through the middle point of AD draw a vertical line ; its
intersections with horizontal lines through A and D determine
the points G and E, so that the two forces GA and DE become
known. The only remaining unknown forces are the vertical
forces .ZTFand FG. Choose a pole O, draw rays to the points
G, A, B, C, D, E, and then the corresponding strings. Through
the points determined by the intersection of og wither, and oe

STRESSES DUE TO WIND PRESSURE.

with ef, draw the string of. The corresponding ray drawn
from O intersects EG in the point F, thus determining EF and
FG. The reactions are now wholly known ; that at the left
support being FA, and that at the right support DF.

D E

(3) For the third case the construction is shown in Fig. 33
(A) and (B), for the two opposite directions of the wind. The
method is identical with that employed in Art. 39. Only one
point of the line of action of the left reaction is known, hence
this is taken as the point of intersection of the corresponding
strings of the funicular polygon. One of these strings can be
drawn at once, since the corresponding ray is known; and the
other is known after the remaining strings have been drawn,
since it must close the polygon. The construction should be
carefully followed through by the student.

The force polygons for the two directions of the wind are
distinguished by the use of O and O' to designate the two poles.

70 GRAPHIC STATICS.

In Fig- 33 (A), ABCDEFGHIA is the force polygon for the
case when the wind is from the right. Notice that the points
A, B, C, D, coincide. This means that the loads AB, BC, CD,
are each zero.

In diagram (B), ABCDEFGHI A is the force polygon fo*
the case when the wind is from the left. The points E, F, G, H,
coincide, because the loads EF, FG, GH, are each zero.

93. Stress Diagrams for Wind Pressure. — When the loads
and reactions due to wind pressure are known, the internal
stresses can be found by drawing a stress diagram, just as in
the case of vertical loads. The construction involves no new
principle, and will be readily made by the student. In Figs.
33 (A) and 33 (B} are shown the diagrams for the two directions
of the wind.

MAXIMUM STRESSES. 7!

The stresses in all members of the truss must be determined
for each direction of the wind. If the truss is symmetrical
with respect to a vertical line, as is usually the case, it may be
that the same stress diagram will apply for both directions of
wind. This will be so if the reactions are assumed to act as
in cases (i) and (2) of the preceding article. In the case repre-
sented in Fig. 33, however, the hinging of one end of the truss
destroys the symmetry of the two stress diagrams, and both
must be drawn in full.

§ 4. Maximum Stresses.

94. General Principles. — For the purpose of designing any
truss member, it is necessary to know the greatest stresses to
which it will be subjected under any possible combination of
loads.

Stresses are combined in accordance with the principle stated
in Art. 78, that the resultant stress in a truss member due to
the combined action of any loads is equal to the algebraic sum
of the stresses due to their separate action.

The method will be illustrated by the solution of an example
with numerical data.

95. Problem — Numerical Data. — Let it be required to design
a wrought iron truss of 40 ft. span, of the form shown in Fig.
34 (PI. I). Let 12 ft. be the distance apart of trusses, and let
the loads be as follows :

Weight of truss, to be assumed in accordance with the
formula of Art. 82 : W=% al (i +^ /). This gives W= 1800 Ibs.
Assuming this to be divided equally among the upper panels,
and that the load for each panel is borne equally by the two
adjacent joints, the load at each of the joints be, cd, de is 450
Ibs. The loads at the end joints may be neglected, being borne
directly at the supports.

Weight of roof. — This depends upon the materials used and
the method of construction, but will be taken as 6 Ibs. per sq. ft.

72 GRAPHIC STATICS.

of roof area, giving 900 Ibs. as the load at each joint. This,
also, is a permanent load. Total permanent load per joint,
1350 Ibs.

Weight of snow. — Taking this as 15 Ibs. per horizontal
square foot, we find 1800 Ibs. as the load at each joint.

Wind pressure. — This is computed from the formula

-A- (Art. 85.)

For this case we put sin a = -£-| = •§• ; /n = 4O Ibs. per sq. ft.;
whence /a=35 Ibs. per sq. ft. (about). This gives upon each
panel of the roof 5250 Ibs. Then with the wind from either
.side, the wind loads on that side would be 2625 Ibs., 5250 Ibs.,
2625 Ibs. respectively.

96. Stress Diagrams. — We are now ready to construct the
stress diagrams.

The truss is shown (PI. I) in Fig. 34 (A). Fig. 34 (B) is the
stress diagram for permanent loads. No diagram for snow
loads is needed, since it would be exactly similar to that for
permanent loads. The snow load at any joint being four-thirds
as great as the permanent load, the stress in any member due
to snow is four-thirds that due to permanent loads.

Fig. 34 (C) shows the stress diagram for the case of wind
blowing from the left. The reactions are assumed to act in
lines parallel to the loads — that is, normal to the roof. With
this assumption, no separate diagram is needed for the case of
wind from the right, since such a diagram would be exactly
symmetrical to Fig. 34 (C). For example, the stress in the
member gh due to the wind blowing from the right is given by
the line GM in Fig. 34 (C).

97. Combination of Stresses. — After the stress diagrams are
•completed for the various kinds of loads, the stresses should
be scaled from the diagrams and entered with proper sign in a
table, as follows :

MAXIMUM STRESSES.

Member.

Permanent
Load.

Snow.

Wind R.

'Wind L.

Max.

-4560

-6080

— 6270

-7925

- 18570

-3760

— 5010

— 6270

-7925

— 16700

- 1080

- 1440

-5250

- 7770

+ 1785

+ 2380

+ 55°

+ 6650

+ 10820

+ 3735

+ 4980

4- 2600

+ 8800

+ 17520

4- 2190

+ 2920

+ 2300

+ 7410

gni

+ 3735

+ 4980

+ SSoo

-f 2600

+ 17520

+ 1785

+ 2380

+ 6650

+ 55°

+ 10820

- 1080

-1440

-5250

- 7770

-3760

— 5010

-7925

— 6270

— 16700

-4560

-6080

-7925

— 6270

- 18570

By combining the results, the maximum stress in each mem-
ber for any possible condition of loading can be determined.
The possible combinations of loading are the following : Perma-
nent load alone ; permanent and snow loads ; permanent load,
and wind from either direction ; permanent and snow loads, and
wind from either direction. The student will readily under-
stand the method of combining the separate results.

The problem here solved relates to a very simple form of
truss. With some forms there may occur a reversal of stress
in certain members, under different conditions of loading.

It is to be noticed that in the table the word maximum is
used in its numerical sense, and has no reference to the algebraic
sign of the stress.

98. Examples. — In Fig. 35 (A) to (F), are shown several
forms of truss for which the student may draw stress diagrams,
assuming loads in accordance with the data given in Arts. 82
to 85. In determining reactions due to wind pressure, the

GRAPHIC STATICS.

three assumptions mentioned in Art. 92 should all be used in
different cases, that the student may become familiar with the
principle of each.

Fig. 35

§ 5. Cases Apparently Indeterminate.

99. Failure of Usual Method. — In attempting to construct
the stress diagram by drawing the force polygon for each joint
in succession, as in the cases thus far treated, a difficulty is
met in certain forms of truss. It may happen that after pro-
ceeding to a certain point it is impossible to select a joint for
which the force polygon can be completely drawn, the number
of unknown forces for every joint being greater than two.

Thus, in the truss shown in Fig. 36, if the stress diagram is
started in the usual way, beginning at the left support, the force
polygons for three joints may be constructed without difficulty,
thus determining the stresses in bl, Ik, Im, cm, mn, nk. But
the force polygon for cdqpnmc cannot be constructed, since
three forces are unknown, — namely, those in dq, qp, pn. And
at the joint knpsk, the stresses in np, ps, and sk are unknown.
The problem, therefore, seems at this point to become indeter-

CASES APPARENTLY INDETERMINATE. 75

minate, since either of the two polygons can be completed in
any number of ways, so far as the known forces determine. It
can be shown, however, that this ambiguity is only apparent.
This may be proved as follows :

Consider the portion of the truss to the left of the broken
line M'N'. It is in equilibrium under the action of eight
forces ; five of these (four loads and a reaction) are known ;
the remaining three are the forces in er, rs, and sk. Now, the
problem of determining these three unknown forces is the same
as that treated in Art. 40. It was there found to be a deter-
minate problem, unless the lines of action of the three unknown
forces intersect in a point or are parallel.

That the problem is determinate may be seen also from the
principle of moments (Art. 51). The eight forces mentioned
being in equilibrium, the sum of their moments is zero for any
origin in their plane. Let the origin be taken at the point of
intersection of the lines of action of two of the unknown forces,
as ert rs. Then from the principle of moments we have (since
the moments of the two forces named are zero) : Algebraic sum
of moments of loads and reaction to left of section + moment
of SK=o. The only unknown quantity in this equation is the
magnitude of SK, which may, therefore, be determined. The
other unknown forces may be found in a similar manner,
the origin of moments being in each case chosen so as to
eliminate two of the three unknown forces.

The whole problem of drawing the stress diagram is now
seen to be determinate. For, as soon as the stress in sk is
known, the force polygon for the joint knpsk contains but two
unknown sides, and can be drawn at once. No further diffi-
culty will be met.

100. Solution of Case of Failure — First Method. — The

reasoning of the preceding article suggests two methods of
treating the so-called ambiguous case. These will now be
described.

76 GRAPHIC STATICS.

The first method is to apply the construction of Art. 40, as
follows : Referring to Fig. 36, consider the equilibrium of the
portion of the truss to the left of the line M'N'. The system
of forces consists of those whose lines of action are ka, ab, be,
cd, de, er, rs, sk. Let them be taken in the order named, and
draw the force polygon for the known forces. (The reaction
ka is supposed to be already determined.) The known part of
the force polygon is KABCDE', the unknown part is to be
marked ERSK. Choose a pole O and draw rays to K, A, B,
C, D, and E ; then draw the corresponding strings of the funic-
ular polygon. Remembering the method of Art. 40, we draw
first oet making it pass through the intersection of er and rs

(the reason for this being that it 'makes the string os also pass
through that point} ; then draw in succession od, oc, ob, oa, ok.
The string ok intersects sk in a point through which os must be
drawn. Hence os must join that point with the starting point
of the polygon. This completes the funicular polygon, except
the string or, which must pass through the intersection of er

CASES APPARENTLY INDETERMINATE.

and rs in some direction not yet known. This string is not
necessary to the solution of the problem.

Since os is now known, OS may be drawn parallel to it from
O ; and by drawing a line from K parallel to the known direc-
tion of KS, the point 5 is determined, and KS becomes
known.

To find ER and RS it is only necessary to draw from 5 a
line parallel to rs, and from E a line parallel to cr ; their inter-
section gives the point R, and the force polygon for the system
of forces considered is complete. The stresses in er, rs, and sk
being now known, the stress diagram may be completely drawn
by the usual method.

It is interesting to notice that the method just described
determines the lines ER, RS, SK, in their proper position in
the complete stress diagram. The determination of ER and
RS by this method is not necessary, since the usual method of
drawing the stress-diagram can be carried out as soon as SK
is known.

The funicular polygon employed in the above construction,
so far as it belongs to the external forces, may coincide with
the corresponding part of the funicular polygon used in deter-
mining the reactions. If this is desired, two points must
be observed: (i) the string oe must be the first drawn, and
must pass through the intersection of er and rs, and (2) the
pole must be so chosen that ok will not be nearly parallel
to ks.

It should also be noticed that the construction of the funic-
ular polygon might begin with the string ok, which should then
be made to pass through the intersection of rs and sk. The
student will be able to carry out this construction without
difficulty.

101. Solution of Case of Failure — Second Method. — It will
now be shown how the apparently ambiguous case can be

78 GRAPHIC STATICS.

treated graphically by the principle of moments. Referring
again to Fig. 36, consider the portion of the truss to the left
of section M'lV. It is acted upon by eight forces, of which
five (the loads and the reaction) are known, and three (whose
lines of action are er, rs, sk) are unknown. The algebraic sum
of the moments of all these forces about any origin must be
zero. Let the origin be taken at the point of intersection of
er and rs, so that the moments of the forces acting in these
lines are both zero ; then the sum of the moments of the five
known forces, plus the moment of the force acting in the line
sk, must equal zero. Now the sum of the moments of the
five known forces may be found by the method of Art. 56.
Through the origin of moments draw a line parallel to the
resultant of the forces named (that is, a vertical line), and let
i equal the length intercepted on it by the strings oe, ok. Then
the required moment is —iH, where //is the pole distance.
(The minus sign is given in accordance with the convention
that left-handed rotation shall be positive.) Let P = unknown
force in line sk, and h its moment-arm. For the purpose of
computing the moment, assume the stress in sk to be a tension ;
then the force P acts toward the right and its moment is posi-
tive, the value being -{-Ph.

Hence, Pk-Hi=o\

P=LH.

From this equation P may be computed. The computation
may be made graphically as follows : Draw (Fig. 36) a triangle
WUV, making VVU=H (force units) and WV=k (linear
units). Lay off WY=i (linear units) and draw YX parallel
to VU. Then WX (force units) represents P. This is readily
seen, since from the two similar triangles we have the propor-
tion — = -, which agrees with the equation above deduced.
H h

The computation is simplified if the pole distance H is taken
equal to as many force units as h is linear units ; or if H

CASES APPARENTLY INDETERMINATE. 79

is some simple multiple of h. For, suppose H=nk\ then

pJa!L = nif If n=Itp = t.

h
The stress in sk is found to be a tension, since P is positive.

Whatever the nature of the stress, it may be assumed a ten-
sion in writing the equation, and the sign of the value found
will show whether the assumption coincides with the fact.

1 02. Other Methods for Case of Failure. — In certain cases
the method of treating the " ambiguous case" may profitably
be varied.

(1) The construction of Art. 100 may be modified as fol-
lows : Determine the resultant of the known external forces
acting on the portion of the truss to one side of the section
M'N'. This resultant is in equilibrium with the three unknown
forces acting in the members er, rs, sk. Hence these forces can
be determined by the special method explained in Art. 42.

The resultant of the five known forces is represented in
magnitude and direction by KE in the force polygon ; and its
line of action passes through the intersection of the strings oey
ok in the funicular polygon. Since this point of intersection is
likely to be inaccessible, the construction of Art. 42 cannot be
conveniently applied. It may be modified by using instead of
KE the two forces KA (the reaction in the line ka) and AE
(the resultant of the four loads, its line of action being deter-
mined by the intersection of the strings oa and oe). First
determine forces in the three lines er, rs, sk, which shall be in
equilibrium with KA ; then make a similar construction for
AE, and combine the results.

(2) It has been proposed to employ the following reasoning :
Remove the members/^ and qr, and insert another represented
by the broken line in Fig. 36. Evidently this does not change
the stress in the member sk, since the forces acting on the
truss to the left of the section M'N' are unchanged. But with
this change the difficulty encountered in constructing the stress
diagram by the usual method is avoided. For when the joint

8o GRAPHIC STATICS.

nmcdqpn is reached, the forces acting there will be all known
except two. Let the stress diagram be drawn in the usual way
until the stress in sk is known. Then restore the original
bracing and repeat the construction, using the value just deter-
mined for SK.

This method is convenient whenever it is applicable. Cases
may, however, arise, in which it will fail. For instance, if
a load is applied at the joint pgrsp, the members pq and qr
cannot both be omitted, and the method cannot be applied. In
such cases, one of the methods explained in the preceding
articles may be applied.

Other methods might be mentioned, but the foregoing dis-
cussion of the case will probably be found sufficient.

103. Failing Case in Other Forms of Truss. — The usual
method of constructing the stress diagram may fail in other
forms of truss, though the one above described is the most
common. In such a case, the problem of finding the stresses
may be really indeterminate, or only apparently so. Whenever
it is possible to divide the truss into two parts by cutting three
members which are not parallel and do not intersect in one
point, the stresses in the three members cut are determinate as
soon as all external forces are known, and can be found by
methods already given. If more than three members are cut,,
the problem of finding the stresses in them is indeterminate,
unless all but three of these stresses are known. By remem-
bering these principles, the determinateness of any given prob-
lem may readily be tested. (See Art. 70.)

§ 6. Three-Hinged Arch.

104. Arched Truss Defined. — If a truss is so supported that
when sustaining vertical loads the reactions at the supports
have horizontal components directed toward the center of the

THREE-HINGED ARCH.

span, the truss becomes an arch. The only kind of arch we
shall here consider is that consisting of two partial trusses
hinged together at the crown, and each hinged at the point of
support.

Such a truss is shown in Fig. 38, in which the two partial
trusses are hinged to the abutments at P and Q, and connected
by a hinge at the point ^?. Since a hinge at the support allows
the reaction of the supporting body upon the truss to take any
direction in the plane of the truss, the directions of the reactions
at P and Q are unknown, as is also that of the force exerted by
either partial truss upon the other at the point R. The problem
of determining the reactions may, therefore, at first sight seem
indeterminate. It will be shown in the next article that it is
in reality determinate, and that the only principles needed in
the solution are such as have been already often applied in the
preceding chapters. The three-hinged arch is indeed a " simple "
structure (Art. 75), since the theory of elasticity is not needed
in the determination of the reactions.

105. Reactions Due to a Single Load. — The method of find-
ing the reactions is most clearly understood by considering the

(JO

(T)

effect of a single load on either partial truss. Let a load be
applied at 5 (Fig. 38) and let all other loads acting on either
portion of the truss (including the weight of the structure) be

.82 GRAPHIC STATICS.

neglected. Call the two partial trusses X and F, and consider
the part Y. The only forces acting upon it are the reaction
exerted by the abutment at Q and the force exerted at R by
the truss X. These two forces, being in equilibrium, must
have the same line of action, which is, therefore, the line QR.
Consider now the body X. The external forces acting upon it
are the load at S, the reaction of the abutment at P, and the
force exerted by Fat the point R. But this last force is equal
and opposite to the force exerted by X upon F, and its line of
action^ is therefore QR. The three forces acting upon the
body X being in equilibrium, their lines of action must meet in
a point ; which point is found by prolonging QR to meet the
line of action of the applied load. Let T be this point, then
PT is the line of action of the reaction at P. The reactions
can now be determined by drawing the triangle of forces. This
triangle is ABC in the figure, AB representing the load at S, BC
the reaction in the line QR (also marked be), and CA the reac-
tion in the line PT (marked also ca). Evidently ABC may be
regarded as the polygon of external forces, either for the partial
truss X, or for the whole structure composed of X and F; and
BC represents either the force exerted by F upon X at R, or
the force exerted upon F by the supporting body at Q.

If, now, the structure be loaded at other points, the reactions
due to each load may be found separately ; the resultant of
all such separate reactions at either support will be the true
reaction at that support when all the loads act together. A
convenient method of applying these principles will be given in
the next article.

1 06. Reactions and Stresses Due to Any Vertical Loads. — In

Fig. 39 is represented a truss consisting of two parts supported
by hinges at P' and Q' and hinged together at R' . Consider
all vertical loads to be applied at the upper joints, their lines of
action being marked in the usual way. We shall first explain
the construction for finding the reactions at the supports ; after

THREE-HINGED ARCH. 83

these are determined, the drawing of the stress diagram will
present no difficulty.

Since we shall sometimes deal with one of the partial trusses,
and sometimes with the two considered as a single body, it will
be well at the outset to specify the external forces acting on
each of these bodies.

Fig*. 30

(i) For the partial truss at the left we have five applied loads,
the reaction at R' (exerted by the other truss), and the reaction
at P! . The force polygon for these forces will be marked as
follows: ABCDEFLA. (The meaning of the letters will be
understood before the force polygon is actually drawn, by
reference to the corresponding letters in the space diagram.)

84 GRAPHIC STATICS.

(2) For the right-hand partial truss the external forces are
five loads and two reactions, and the force polygon will be
marked thus : FGHIJKLF. (Notice that FL and LFare equal
and opposite forces, being the " action and reaction" between
the two trusses at R' .)

(3) For the combined structure the external forces are the
ten loads and the reactions at P1 and Q' . (The action and
reaction at R! become now internal forces.) The force polygon
will be marked thus : ABCDEFGHIJKLA.

Begin the construction by drawing the force polygon for the
ten loads on the whole structure, lettering it as just indicated.
Choose a pole O, draw the rays, and then the funicular polygon
as far as possible. Now consider the right partial truss as
unloaded. The resultant of the remaining five loads is repre-
sented in magnitude and direction by AF\ its line of action
must pass through the intersection of oa and of, and is therefore
the line marked of. Now, reasoning as in the preceding article,
we see that the reaction at Q' must act in the line Q'R'. Let
Q'R' intersect of in Tf, then P' T' is the line of action of
the reaction at P'. Hence the complete force polygon for the
whole truss, when the right half is unloaded, may be found by
drawing from Fa. line parallel to Q' R' , and from A a line parallel
to P' T', prolonging them till they intersect at L' . The reaction
at P' is L'A, and that at Q' is FL1. (The line of action of the
latter is marked fl'.)

Next, consider the left half to be without loads, the other
half being loaded. The resultant of the five loads now acting is
FK, its line of action fk being drawn through the point of
intersection of of and ok. The reactions at P' and Q' for the
present case have lines of action P'Rf and Q'T", found just
as in the first case of loading. These reactions are therefore
determined in magnitude and direction by drawing from K a
line parallel to Q'T" and from F a line parallel to P'R', pro-
longing them till they intersect at L". The complete force
polygon for this case of loading is therefore FGHIJKL"F.

THREE-HINGED ARCH. 85

Consider now that both parts of the truss are loaded. The
reaction at P1 is the resultant of the two partial reactions L'A,
L"F, and the reaction at Q' is the resultant of the two partial
reactions FL' , KL". From L" and L' draw lines parallel
respectively to FL' and FL", intersecting in L. Then L" L is
equal and parallel to FL' , and LL' is equal and parallel to L"F.
Hence KL and LA represent the resultant reactions at Q' and
P' respectively. This completes the polygon of external forces
for the whole truss, as well as that for each partial truss.

The stress diagram can now be drawn in the usual way,
beginning at the point P' . The diagram for one partial truss
is shown in Fig. 39 (B).

If loads are applied at the lower joints of the trusses, the
reactions due to these may be found in the same manner as for
the upper loads. But before beginning the determination of
the stresses, the polygon must be drawn for the external forces
taken in order around the truss. (See Art. 90.)

107. Case of Symmetrical Loading. — If the two half trusses
are exactly similar and symmetrically loaded, the determination
of reactions and stresses is much simplified.

(1) As regards the reactions, symmetry shows that the forces
exerted by the two trusses upon each other at R' are horizontal.
Hence, referring to Fig. 39, and considering either half-truss,
as that to the left, the line of action of the reaction at P' may
be found by drawing a horizontal line through R1 and prolong-
ing it to intersect of, the line of action of the resultant of all
loads on the left truss ; the line joining this point of intersection
with P' is the required line of action of the reaction at the left
abutment. The two reactions are now determined by drawing
from Fa horizontal line and from A a line parallel to the line
just determined, and prolonging them till they intersect.

(2) As to the stresses, only one partial truss need be consid-
ered, since the stress diagrams for the two portions will be
symmetrical figures.

GRAPHIC STATICS.

These principles might have been employed in the case dis-
cussed in the preceding article ; but the method there used is
applicable to cases in which either the trusses or the loads are
unsymmetrical.

108. Wind Pressure Diagram. — The diagrams for wind
pressure will present no difficulty. The determination of the
reactions will, indeed, be simpler than in the preceding case,
since only one partial truss will be loaded at any one time, and
the line of action of one reaction is therefore known at the
outset. The construction is shown in Fig. 40.

Fig. 4O.

In computing wind pressure loads, it will be assumed that
each joint sustains half the pressure coming upon each of the
two adjacent panels. It will be sufficiently correct in comput-
ing the pressure at any joint, to assume the slope of the roof

THREE-HINGED ARCH. 8/

as that of the tangent to the roof curve at the joint in ques-
tion ; and the direction of the wind load may be taken as that
of the normal to the roof curve at the joint. The force poly-
gon for the wind loads is therefore not a straight line when the
roof is curved. In Fig. 40, this polygon is FGHIJK. The
reactions are to be marked KL, LF. (Since there are no loads
on the other half-truss, the points ABCDEF will coincide.)
Choosing a pole O, draw the funicular polygon as shown, and
determine the line of action of the resultant of all the wind
loads. This line of action //£ is drawn parallel to FK, through
the point of intersection of the strings of and ok. Prolong//^
to intersect P'R' produced at T" ; then Q'T" is the line of
action of the reaction at Q'. The force polygon may now be
completed by drawing KL parallel to Q'T" and LF parallel to
P'R'. The reactions KL and LF being thus determined, the
stress diagram can be drawn without difficulty.

The stress diagram is drawn for both partial trusses, with the
result shown in the figure. If the two trusses are symmetrical,
the diagram for the other direction of the wind need not be
drawn ; the stresses for this case being found from the diagram
already drawn. If, however, the partial trusses are dissimilar, a
second wind-diagram must be drawn.

It is not necessary to draw separate space diagrams for verti-
cal loads and wind-forces. The constructions of Figs. 39 and
40 have been here kept wholly separate, in order that the
explanation may be more easily followed.

109. Check by Method of Sections. — In case of a truss of
long span, especially when the members have many different
directions and are short compared with the whole length of the
span, the small errors made in drawing the stress diagram are
likely to accumulate so much as to vitiate the results. Thus,
in Fig. 39, if, in drawing the stress diagram, we begin at P' and
pass from joint to joint, there is no check upon the correctness
of the work until the point R' is reached. At that point we

88 GRAPHIC STATICS.

have a second determination of the reaction exerted by each
half-truss upon the other ; and it is quite likely that the two
values found will not agree.

By a method like that employed in Art. 100 for the "inde-
terminate " form of truss, we may avoid the necessity of mak-
ing so long a construction before checking the results. In
Fig. 39 take a section cutting the three members cq, qr, rL
and apply the principles of equilibrium to the body at the
left of the section. The forces acting on this body are LA,
AB, BC, CQ, QR, RL. Choose a pole O' and draw the funic-
ular polygon for this system, making the string o'c pass through
the point of intersection of cq and qr. Draw successively the
strings o'c, o'b, o'a, <?'/, prolonging the last to intersect Ir.
Through the point thus determined, draw o'r, which must also
pass through the point of intersection of cq and qr. As soon as
o*r is known, the corresponding ray O'R can be drawn in the
force diagram, and then by drawing from L a line parallel to
Ir, the point R is determined. We may now close the force
polygon, since the directions of the two remaining forces (cq
and qr) are known.

The stresses in the three members cut being now known, we
have a check on the correctness of the construction of the
stress diagram, as soon as these members are reached in the
process.

This method will be found of great use, not only for this
form of truss, but for any truss containing many members.

§ 7. Counterbracing.

no. Reversal of Stress. — If the loads supported by a truss
are fixed in position and unchanging in amount, the stress in
.any member remains constant in magnitude and kind. But in
most cases such are not the conditions, and it may happen that
under different combinations of loading, the stress in a certain
member is sometimes tension and sometimes compression. It

COUNTERBRACING.

is often thought desirable to prevent such changes of stress,
the design of the members and their connections being thereby
simplified. To accomplish this is the object of counterbmcing.

in. Counter bracing. — Consider a truss such as the one
shown in Fig. 41, subjected to vertical loads and to wind pres-
sure from either side. A diagonal member such as xy may
sustain tension under vertical loads alone, or with the wind
from the left ; while with the wind from the right it may sustain
compression.

Now suppose xy removed, and a member represented by the
broken line xy' introduced. It may easily be shown that any

,'M

system of loading which would cause compression in xy will
cause tension in this new member ; and vice versa. For, divide
the truss by a section MN cutting xy and two chord mem-
bers as shown, and let L be the point of intersection of the
two chord members produced. The kind of stress in xy or
the other member (whichever is assumed to be present) may
be determined by considering the system of forces acting on
•one portion of the truss, as that to the left of the section MN.
Let the principle of moments be applied to this system, the
origin being taken at L. If the external forces acting on the
portion of the truss considered be such as to tend to cause
right-handed rotation about L, the stress in xy must be com-
pression in order to resist this tendency ; while, if xy be replaced
by xy', a tension must exist in that member to resist right-
handed rotation about L. Similarly, a tendency of the external
forces to produce left-handed rotation about L would be resisted
by a tension in xy ; or by a compression in the member xy' .
If the two members act at the same time, the stresses in

90 GRAPHIC STATICS.

them will be indeterminate. These stresses may, however, be
made determinate by the following device :

Let the member xy be so constructed that it cannot sustain
compression. Then, whenever the external forces are such as
to tend to throw compression upon it, it ceases to act as a
truss-member, and the member xy' receives a tension which is
determinate.

If the member xy1 be constructed in the same way, any
tendency to throw compression upon it causes it immediately
to cease to act, and puts upon the member xy a tension instead.

A member such as xy\ constructed in the manner mentioned,
is called a counterbrace.

112. Determination of Stresses with Counterbracing. — The

use of counterbracing adds somewhat to the labor of deter-
mining the maximum stresses, since the members actually under
stress are not always the same. The method of treating such
cases will be illustrated in the next article by the solution of an
example ; but first the main steps in the process may be outlined
as follows :

(a) Construct separate stress diagrams for vertical loads and
for wind in each direction, assuming the diagonals in all panels
to slope the same way.

(b) Determine by comparison of these diagrams in which of
the diagonal members the resultant stress is ever liable to be a
compression. Draw in counters to all such members.

(c) With these counterbraces substituted for the original
members, either draw new stress diagrams, or make the neces-
sary additions to those already drawn. If the latter method is
adopted, the added lines should be inked in a different color
from that employed originally. (In some cases, this construc-
tion may be unnecessary on account of symmetry, as will be
illustrated in the next article.)

(d) Combine the separate stresses for maxima in the usual
way.

COUNTERBRACING. 9!

113. Example. — In Fig. 42 (PI. II) are given stress diagrams
for a "bow-string" roof-truss in which counterbracing is em-
ployed. At (A) is shown the truss or space diagram. The
span is 48 ft. ; rise of top chord, 16 ft. ; rise of bottom chord,
8 ft. The chords are arcs of parabolas. The whole truss is
divided into six panels by equidistant vertical members. The
distance apart of trusses is taken as 12 ft.

Assume the weight of the truss at 2400 Ibs., and that of the
roof at 3600 Ibs. ; this makes the total permanent load 1000 Ibs.
per panel. Take 800 Ibs. as the load at each upper joint, and
200 Ibs. as the load at each lower joint.

The snow load, computed at 15 Ibs. per horizontal square
foot, is about 1440 Ibs. per panel.

Wind pressure is to be computed from the formula of Art. 85.

We now proceed to apply the method outlined in the preced-
ing article.

(a) Assuming one set of diagonals present, we construct the
stress diagrams for the various kinds of loads.

Diagram for permanent loads. — This is shown at (B) Fig.
42, which needs little explanation. It will be noticed that the
stress in every diagonal member is zero. This will always be
the case if the chords are parabolic and the vertical loads are
equal and spaced at equal horizontal distances.

Diagram for snow loads. — Fig. 42 (C) is the stress diagram
for snow loads. In this case, also, the stresses in the diagonals
are all zero.

Wind diagrams. — At (D) and (E) are shown the diagrams
for the two directions of the wind. The only thing needing
special mention is the method used in laying off the force-
polygon for the wind loads. We first compute the normal wind
pressure on each panel by the formula of Art. 85. We thus
find, when the wind is from the right, the following total pres-
sures, taking /n = 40 Ibs. per sq. ft. :

On panel d, 1650 Ibs. On panel e> 3870 Ibs. On panel/, 5480 Ibs.

GRAPHIC STATICS.

In Fig. 42 (D) these are laid off successively in their proper
directions to the assumed scale. Thus CD1, D'E', E'G repre-
sent respectively 1650 Ibs. normal to dq, 3870 Ibs. normal to es,
and 5480 Ibs. normal to //. Now each of these loads is to be
equally divided between the two adjacent joints. Bisect CD' at
D, D'E' at E, and E'G at F\ then CD, DE, EE, EG represent
the loads at the joints cd, de, ef, and fg respectively. The
"load line" is therefore CDEEG.

The reactions are assumed to be parallel to the resultant
load. With this explanation the figures (D) and (E) will be
readily understood.

(b) Comparison of results. — The stresses due to permanent
loads, wind right, and wind left, are shown in tabular form for
convenience of comparison.

Member.

Perm. Load.

Snow Load.

Wind R.

Wind L.

Max.

- 6660

-9680

- 4250

- 14070

- 16340

-5350

-7780

- 4880

- 9650

- I3I30

-455°

— 6640

- 6380

— 6200

— 11190

-4550

— 6640

- 95°°

- 4250

— 14050*

-5350

-7780

- 15030

- 345°

- 20380*

-6660

-9680

— 14070

- 43oo

- 20730*

+ 5080

+ 7400

+ 780

4- 13500

+ 12480

kh&

+ 4680

+ 6820

+ 700

+ 12450

+ 11500

mh±

+ 4470

+ 6520

+ 1950

+ 735°

+ 10990

Ph*

+ 447°

+ 6520

+ 4100

+ 10990*

rh.

+ 4680

+ 6820

+ 7680

+ 2050

+ 12360*

th,

+ 5080

+ 7400

+ 13500

+ 800

+ 18580*

+ 1180

+ 1440

+ 15°

+ 2650

4- 2620

+ 1180

+ 1440

- 280

+ 4100

+ 2620

+ 1180

+ 1440

- 95°

+ 3800

+ 2620*

+ 1180

+ 1440

— 1650

+ 2525

/ + 2620* ]
I - 470* /

+ 1180

+ 1440

- 135°

+ 1250

f + 2620* |

1 * (

I — 170* '

+ 1300

- 4600

+ 1300*

+ 2700

— 4100

+ 2700*

+ 4850

- 315°

+ 4850*

+ 7080

- 195°

+ 7080*

COUNTERBRACING. 93

It is seen that the diagonals shown are all in tension when
the wind is from the right, and all in compression when the
wind is from the left. Therefore counters are needed in all
panels, and the counters will all come in action whenever the
wind is from the left.

(c) Stresses in counterbraces. — It is evident from symmetry
that no new diagrams are needed to determine the stresses in
the counterbraces. In fact, the counterbrace in each panel is
situated symmetrically to the main diagonal in another panel,
and is subject to exactly equal stresses.

(d) Combination for maxima. — In combining the results for
the greatest stresses in the various members, it will be assumed
that the greatest snow load and the greatest wind load can
never act simultaneously. For each member, therefore, the
stress due to permanent load is to be added to the snow stress
or the wind stress, whichever is greater. Again, in combining
the tabulated results, we consider only the columns headed per-
manent load, snow load, and wind right ; since whenever the
wind is from the left, the other system of diagonals is in action.
This gives the results entered in the last column.

We now notice the following facts :

(1) The results given for the diagonal members are the true
maximum stresses.

(2) The stress found for each diagonal applies also to the
symmetrically situated counterbrace.

(3) For any other member, we are to choose between the
maximum found for that member and the value found for the
symmetrically situated member. Thus, —20730 is the true
maximum stress for both aj and//; etc. (The numbers denot-
ing true maximum stresses are marked with a (*) in the table.)

It is seen that the verticals, with one exception, may sustain
a reversal of stress. Thus,/£ and st must be designed for a
tension of 2620 Ibs., and also for a compression of 170 Ibs. ;
while Im and qr are each liable to 2620 Ibs. tension and to 470
Ibs. compression.

CHAPTER VI. SIMPLE BEAMS.

§ i. General Principles.

1 14. Classification of Beams. — A beam has been defined in
Art. 79. Beams may be treated in two main classes, the basis
of classification being that described in Art. 75. These two
classes will be called simple and non-simple beams respectively.
The present chapter deals only with simple beams, the defini-
tion of which may be stated as follows :

A simple beam is one so supported that it may be regarded
as a rigid body in determining the reactions.

A simple beam may rest on two supports at the ends ; or it
may overhang one or both supports.

A cantilever is any beam projecting beyond its supports.
Such a beam may be either simple or non-simple.

A continuous beam is one resting on more than two supports.
Such a beam is non-simple.

A beam may be supported in several ways. It is simply
supported at a point when it rests against the support so that
the reaction has a fixed direction. It is constrained at a point
if so held that the tangent to the axis of the beam at that
point must maintain a fixed direction. If hinged at a support,
the reaction may have any direction. We shall deal mainly
with the case of simple support.

In what follows it will be assumed that the beam rests in a
horizontal position, since this is the usual case.

115. External Shear, Resisting Shear, and Shearing Stress.

- The external shear at any section of a beam is the algebraic

GENERAL PRINCIPLES. 95

sum of the external vertical forces acting on the portion of the
beam to the left of the section.

The resisting shear at any section is the algebraic sum of the
internal vertical forces in the section acting on the portion of
the beam to the left, and exerted by the portion to the right
of the section.

The shearing stress at any section is the stress which con-
sists of the internal vertical forces in the section, exerted by
the two portions of the beam upon each other. It consists of
the resisting shear and the reaction to it. (See Art. 63.)

Let AB (Fig. 43) be a beam in equilibrium under the action
of any external forces. At any point in its length, as C, con-
ceive a plane to be passed perpen-
dicular to the axis of the beam, and
consider the portion AC, to the left
of the section. The principles of equilibrium apply to the body
AC, and the external forces acting upon it include, besides
those forces to the left of C that are external to the whole bar,
certain forces acting across the section at C that are internal to
AB, but external to AC. (Art. 61.) These latter forces com-
prise that constituent of the internal stress between AC and
CB which is exerted by CB upon AC.

Represent by Fthe algebraic sum of the resolved parts in
the vertical direction of all forces acting on AB to the left of
the section at C, upward forces being called positive. V is the
external shear at the given section as above denned.

Since the body AC is in equilibrium, condition (i), Art. 58,
requires that the algebraic sum of the resolved parts in the
vertical direction of all forces acting on it must equal zero.
Hence the forces acting on AC m the section at C must have a
vertical component equal to — V. This vertical component is
called the resisting shear in the given section. This resisting
shear is one of the forces of a stress of which the other is an
equal and opposite force exerted by A C upon CB. This stress
is called the shearing stress in the section, and is called positive

96 GRAPHIC STATICS.

when it resists a tendency of AC to move upward, and of CB
to move downward.

1 1 6. Bending Moment, Resisting Moment, and Stress Moment,

— The bending moment at any section of a beam is the algebraic
sum of the moments of all the external forces acting on the
portion of the beam to the left of the section ; the origin of
moments being taken in the section.

The resisting moment at any section is the algebraic sum of
the moments of the internal forces in the section acting on the
portion of the beam to the left, and exerted by the portion to
the right of the section ; the origin of moments being the same
as for bending moment.

The stress moment or moment of internal stress at any section
consists of the two equal and opposite moments of the forces
exerted across the section by the two portions of the beam
upon each other.

Referring again to Fig. 43, let us analyze further the forces in
the section at C. Applying to the body AC the second condition
of equilibrium ((2) of Art. 58), and taking an origin at a point
in the section, we see that the algebraic sum of the moments
of all the external forces acting on the beam to the left of the
section plus the sum of the moments of the internal forces act-
ing on AC in the section must equal zero. The former sum is
denned as the bending moment at the given section. Represent
it by M. The latter sum is denned as the resisting moment at
the section, and must be equal to —M, by the above principle.

We have thus far referred only to the internal forces exerted
by CB upon AC; but evidently the equal and opposite forces
exerted by AC upon CB have a moment numerically equal to
M. The equal and opposite moments of the equal and opposite
forces of the stress in the section together constitute the stress-
moment in the section.

If the external forces applied to the beam are all vertical, the
value of M will be the same at whatever point of the section

GENERAL PRINCIPLES.

the origin is taken ; since the arm of each force will be the same
for all origins in the same vertical line. If the loads and reactions
are not all vertical, the value of M will generally depend upon
what point in the section is taken as the origin of moments.

117. Curves of Shear and Bending Moment. — The curve of
shear for a beam is a curve whose abscissas are parallel to the
axis of the beam, and whose ordinate at any point represents
the external shear at the corresponding section of the beam.

Let AB (Fig. 44) represent a beam loaded in any manner, and
let A'B' be taken parallel to AB. At every point of A'£'
suppose an ordinate drawn whose length shall represent the
external shear at the corresponding B

section of AB. The line ab> join-

is'

ing the extremities of all these ordi- a
nates, is the shear curve. Positive A'
values of the shear may be repre- ^
sented by ordinates drawn upward
from A'B', and negative values by
ordinates drawn downward. (Instead of drawing A'B' parallel
to the beam, it may be any other straight line whose extremi-
ties are in vertical lines through A and B.}

The curve of bending moment for a beam is a curve whose
abscissas are parallel to the axis of the beam, and whose ordinate
at any point represents the bending moment at the correspond-
ing section of the beam. Thus in Fig. 44, A"C"I>" may repre-
sent the bending moment curve for the beam AB. (Evidently
A"£>" might be inclined to the direction of AB, without destroy-
ing the meaning of the curve.) Positive and negative values of
the bending moment will be distinguished by drawing the
ordinates representing the former upward and those represent-
ing the latter downward from the line of reference A"B".

1 1 8. Moment Curve a Funicular Polygon. — If the loads and
reactions upon the beam are all vertical, every funicular poly-
gon for these forces taken consecutively along the beam, is a

GRAPHIC STATICS.

curve of bending moments. Thus, let MN (Fig. 45) represent
a beam under vertical loads, supported at the ends by vertical
reactions. Draw a funicular polygon for the loads and reactions
as shown.

Fig.

Now, by definition, the bending moment at any section is
equal to the moment of the resultant of all external forces act-
ing on the beam to the left of the section, the origin of
moments being taken in the section. By Art. 56, this moment
can be found by drawing through the section a vertical line and
finding the distance intercepted on it by the two strings corre-
sponding to the resultant mentioned ; the product of this inter-
cept by the pole distance is the required moment. Hence, if
oe (Fig. 45) is taken as axis of abscissas, the broken line made
up of oa, ob, oc, od, is a " curve of bending moments."

119. Design of Beams. — The principles involved in the
design of beams will not be here fully discussed. Every prob-
lem in design involves the determination of shears and bend-
ing moments throughout the beam ; and the graphic methods
of determining these will alone be considered in the following
articles.

§ 2. Beam Sustaining Fixed Loads.

120. Shear Curve for Beam Supported at Ends. — Let MN
(Fig. 45) represent a beam supported at the ends and sustain-
ing loads as shown. Draw the force polygon ABCD, and with
pole O draw the funicular polygon. The closing line is oe

BEAM SUSTAINING FIXED LOADS.

(marked also M"N"), and OE drawn parallel to M"N" fixes E,
thus determining DE> EA, the reactions at the supports.

Take M'N' as the axis of abscissas for the shear curve.

The shear at every section can be at once taken from the
force polygon. For, remembering the definition of external
shear (Art. 115) we have :

Shear at any section between M and P is EA (positive).

Shear at any section between P and Q is EB (positive).

Shear at any section between Q and R is EC (negative).

Shear at any section between R and N is ED (negative).

Hence the shear curve is the broken line drawn in the
figure.

121. Moment Curve for Beam Supported at Ends. — As in

Art. 1 1 8, it is seen that the funicular polygon already drawn
(Fig. 45) is a bending moment curve for the given forces. For
the bending moment at any section of the beam is equal to the
corresponding ordinate of this polygon, multiplied by the pole
distance.

For simplicity, it will be well always to choose the pole so
that the pole distance represents some simple number of force-
units.

The sign of the bending moment is readily seen to be nega-
tive everywhere, according to the convention already adopted
(Art. 47).

122. Shear and Moment Curves for Overhanging Beam. —

Consider a beam such as shown in Fig. 46, supported at Q and
T, and sustaining loads at M, P, R, S, and N.

This case may be treated just as the preceding, care being
exercised to take all the external forces (loads and reactions) in
order around the beam.

The construction is shown in Fig. 46. First, the reactions at
Q and T are found as in the preceding case, by drawing the
funicular polygon, finding the closing line, and drawing OG
parallel to it. The reactions are FG and GA. Then the value

100

GRAPHIC STATICS.

of the external shear can be found for any section from the
definition, and is always given in magnitude and sign by a cer-
tain portion of the force polygon ABCDEFGA. The resulting
curve is shown in the figure, M'N' being the line of reference.

c\d die

e\f

p] 70 R js |T N

a\g g\f

Fig. 46

Similarly, from the definition of bending moment, and the
principle of Art. 56, the bending moment at any section is
equal to the ordinate of the funicular polygon multiplied by the
pole distance. The polygon is shaded in the figure to indicate
the ordinates in question. It is seen that the bending moment
is positive at every section of the beam.

123. Distributed Loads. — In all the cases thus far discussed,
the loads applied to the beam have been considered as concen-
trated at a finite number of points. That _is, it has been
assumed that a finite load is applied to the beam at a point.
Such a condition cannot strictly be realized, every load being
in fact distributed along a small length of the beam. If the
length along which a load is distributed is very small, no im-
portant error results from considering it as applied at a point.

In certain cases a beam may have to sustain a load which is
distributed over a considerable part of its length, or over the
whole length. Such a load may be treated graphically with
sufficient correctness by dividing the length into parts, and

BEAM SUSTAINING MOVING LOADS. IOI

.assuming the whole load on any part to be concentrated at a
point. The smaller these parts, the more nearly correct will
the results be.

It may be remarked by way of comparison that while algebraic
methods are most readily applicable to the case of distributed
loads, the reverse is true of graphic methods, which are most
easily applied in the very cases in which analytic methods
become most perplexing.
/

124. Design of Beam with Fixed Loads. — The above exam-
ples are sufficient to explain the method of treating any simple
beam under fixed loads. Under such loading the shear and
bending moment at each section of the beam remain unchanged
in value, and no further discussion is necessary as a preliminary
to the design of the beam. We proceed next to the case of
beams with moving loads.

§ 3. Beam Sustaining Moving Loads.

125. Curves of Maximum Shear and Moment. — When a beam
sustains moving loads, the shear and moment at any section do
not remain constant for all positions of the loads. In such a
case it is the greatest shear or moment in each section that is
to be used in designing the beam.

A curve of maximum shear is a curve of which the ordinate
at each point represents the greatest possible value of the shear
at the corresponding section of the beam for any position of the
loads.

j\ curve of maximum moment is a curve whose ordinate at
each point represents the greatest possible value of the bending
moment at the corresponding section of the beam for any posi-
tion of the loads.

In the following articles will be explained a method of deter-
mining any number of points of the curves just defined, in the
case of a simple beam supported at the ends. The moving

102 GRAPHIC STATICS.

load will be taken to consist of a series of concentrated loads
with lines of actions at fixed distances apart. An example of
such a load is the weight of a locomotive and train ; the
points of application of the loads being always under the several
wheels.

On PL III is represented a load-series consisting of two loco-
motives followed by a train whose weight is assumed to be
uniformly distributed. The numbers given represent half the
total weight of the train, being the weight borne by each of the
two beams or trusses which sustain a single-track railroad.

126. Position of Loads causing Greatest Shear at a Given
Section. — In order to determine the position of a given series
of loads which causes the greatest positive shear at a given
section of a beam, consider first the way in which the shear due
to a single load varies as the position of the load changes.

Let A^B^ (PL III) represent the beam, and F1 the section
under consideration. A load in any~position between F1 and B±
causes at Fl a positive shear equal in magnitude to the reaction
at A^ ; a load between Al and Fl causes at F1 a negative shear
equal to the reaction at Bv

Let AlBl = /, A^ = /p F1^1 = /2. A load P on A^Fly dis-

tant x from Alt causes at Fl a shear -- ; as the load moves

from A1 to F1 this shear varies from o to - — 1. A load P on

F±Blt distant x from Blt causes at F^ a shear — ; as the load

pj
moves from B^ to F^ this shear varies from o to — ?. These

results may be represented graphically as follows :

Influence line. — From a given reference line A^B^ (PL III
(£"))» let ordinates be erected such that the ordinate at any point
y2 represents (to a convenient scale) the shear at F1 due to a
unit load atyr The line joining the extremities of these ordi-
nates is called an "influence line" for the shear in question.

BEAM SUSTAINING MOVING LOADS. 103

The above values of the shear due to a load P in any position
show that for the portion of the beam AlFl the influence line is
a straight line A2F5', F2F3f being equal in magnitude to -j, and

being laid off negatively ; while for the portion F1B1 the influ-
ence line is the straight line B2F3, F2F3 being equal to -2, and

being laid off positively. The two lines A2F3f and F3£2 are seen
to be parallel.

The influence line shows at a glance the effect of a load in
any position in producing shear at F^. Thus, a unit load at J^
causes a positive shear equal to J%J& and a load P at Jl causes
a positive shear equal to P xj^fs-

The resultant shear at Fl due to any number of loads may be
found by multiplying each load by the corresponding ordinate
of the influence line and taking the algebraic sum of the
products.

Consider, now, the effect of any series of loads brought on
the beam from the right. Every load causes positive shear at
Fl until the foremost load reaches the section, and the effect of
each load increases as it approaches F±. As soon as any load
has passed the section, its effect is to cause negative shear, this
effect decreasing as the load recedes from the section.

It thus appears that, in order that the shear at a given section
may have its greatest positive value, (a) there should be little or
no load to the left of the section, (b) the portion of the beam to
the right of the section should be fully loaded, and (c) the loads
at the right should be as near the section as possible.

These rules are not sufficiently definite to serve as an exact
guide. Suppose the first load to be just at the right of Fv and
consider the effect of advancing the whole series of loads. As
the first load passes the section the shear diminishes by an
amount equal to that load. But as the loads advance the effect
of every load in producing reaction at Al increases (thus increas-
ing the shear at F^, while no further decrease occurs until the
second load passes the section. The net result of bringing the

IO4

GRAPHIC STATICS.

second load up to the section may or may not be to increase
the shear over the value it has when the first load is just at
the right of Fv It is thus uncertain, without computation,
whether the first load or some succeeding load should be brought
up to the section in order to cause the greatest positive shear.

Instead of resorting to trials to determine for which position
the shear is greatest, we may apply a simple rule which will
now be deduced.

Let Pl = magnitude of foremost load, P2 = magnitude of sec-
ond load, etc., W being the total load on the beam. Let / =
total span, and m = distance between Pl and P2. Let x = dis-
tance from /?]_ to the center of gravity of J^when Pl is at the
given section. Let us compute the shear when Pl is just at
the right of the section, and then determine the effect of moving
all loads to the left, until the second load comes to the section.

When P1 is at the right of the section, the shear is equal to
the reaction at Alt say R. Then (calling F the external shear)
we have

If now the loads be moved until P2 comes to a point just
at the right of the section, the reaction due to W becomes

and the shear at the section becomes

The increase of the shear is, therefore,

W(x+m) r> Wx Wm p
_____ />!-—= — P,.

This increase is plus or minus according as — ^ is greater or

P /

less than Pl ; that is, as W is greater or less than — L. Hence

the following rule :

BEAM SUSTAINING MOVING LOADS. 105

The maximum, positive shear in any section of the beam occurs

when the foremost load is infinitely near the section, provided W

PI PI

is not greater than — L . If W is greater than — i-, the greatest
m m

shear will occur when some succeeding load is at the section.

In the above discussion it has been assumed that in bringing
P2 up to the section no additional loads are brought upon the
beam. If this assumption is not true, let W be the new load
brought on the beam, and x1 the distance of its center of
gravity from the right support when Pz is at the section. Then
the shear corresponding to this position is

and the increase of shear due to the assumed change in the
position of the load is

Wm.W'x1 n

~T ~T "*

Hence, in the statement of the above rule, we have only to

Wr'

substitute Wm+W'x' for Wm\ or, W+^-^- for W. The

additional load W may be neglected except when the condition

P I

W= — — is nearly satisfied ; for the term W'x1 will always be
m

small.

The application of this rule is very easy, and will save much
labor in the graphic construction of the shear curve.

If it is found that the first load should be past the section
for the position of greatest shear, we may determine whether
the second or the third should be at the section by an exactly
similar method. We have only to apply the above rule, substi-
tuting P2 for Plt and for m the distance between P2 and PB.

127. Determination of Greatest Shear. — Having determined
what position of the load-series causes the greatest shear at a
given section, the value of this shear may be determined by

106 GRAPHIC STATICS.

the method already explained in the discussion of fixed loads
(Art. 120).

Draw in succession the lines of action of the loads at their
proper distances apart (1-2, 2-3, . . ., PL III). Draw the force
polygon * 1-2-3- • • •> choose a pole O, and construct a funicular
polygon for the series of loads. The same lines of action may
be used for all positions of the load-series, for instead of moving
the loads in the desired direction, the loads may be regarded as
fixed and the beam moved in the opposite direction.

At (5), PI. Ill, is shown a beam A^Blt of 64 ft. span, in such
position that the shear at the section Flt 16 ft. from Av has its
greatest possible value. The second load is at the section
instead of the first, for applying the test we have P1 = 8ooo Ibs.,

P /

7=64 ft., m = S.i ft., _i-= 63200 Ibs. ; while the total load on
m

the beam is 104,000 Ibs. when the first load is at the section.

The beam being in the position shown at (5), vertical lines
drawn through A1 and Bl intersecting the strings 01 and 02' at
A and B, determine AB, the closing string of the funicular
polygon for the system of forces acting upon the beam in the
assumed position. If OK is the ray drawn parallel to ABy the
reactions at A1 and Bl are Ki and 2' ' K respectively. The load
2-3 being just at the right of the section Fv the shear at that
section is equal to the reaction at A1 minus the load 1-2 ; its
value is therefore K2 in the force polygon.

The diagram of maximum positive shear is shown at (A),
PL III, the ordinates representing greatest shear due to the
moving loads being laid off upward from A-J3V Thus the value
just determined is laid off from the point Fv The entire curve
is shown at (A), but the construction is given only for the
section Fv

The foregoing construction has referred only to moving loads.
The actual greatest shear at any section is found by combining

* The lines representing the loads are for convenience drawn upward instead of down-
ward. This does not, of course, affect the construction.

BEAM SUSTAINING MOVING LOADS. 107

the maximum live-load shear with the shear due to permanent
loads. Shears due to the dead loads are represented at (A),
PI. Ill, being determined as follows.

128. Shear Curve for Combined Fixed and Moving Loads. -

Let the beam sustain a fixed load of 25000 Ibs. uniformly dis-
tributed along the beam. The shear close to the left support
due to this load is equal to the reaction, or 12500 Ibs.; and
decreases as we pass to the right by ^^- Ibs. for each foot.
At the middle of the beam the shear is zero, and at the right
support it is — 12500 Ibs. Hence the shear curve is a straight
line, and may be drawn as follows : From A^ lay off an ordinate
downward representing 12500 Ibs., and from Bl an ordinate
upward representing 12500 Ibs.; the straight line joining the
extremities of these ordinates is the shear curve for the fixed
load. Positive shears are laid off downward and negative
shears upward for the reason that, if this be done, the greatest
positive shear at any point due to fixed and moving loads is
represented by the total ordinate measured between the shear
curves for fixed loads and for moving loads. It is seen that
at a certain point somewhere to the right of the center of the
beam this greatest shear is zero, and for all sections to the
right of this point it is negative. This point is determined by
the intersection of the two shear curves.

129. Approximate Position of Loads causing Greatest Bending
Moment at a Given Section. — Let it be required to determine the
greatest bending moment at the section Fl of "the beam A^B^
shown on PI. III. Let A1F1 = !1, F^B^l^ AlBl = ll-\-l^ = l.
Consider the effect of a single load in any position.

The bending moment at Fl due to a loadP on A^F^, distant x

PI r
from Av is — j— . For a loadP on F^Blt distant x from Blt the

PI x
bending moment at F^ is — ~- The variation of each of these

values as the position of the load changes may be represented
graphically as follows :

108 GRAPHIC STATICS.

Influence line. — From A^B^ ((£>), PL III), let ordinates be
erected such that the length of the ordinate at any point repre-
sents, on a convenient scale, the bending moment at Fv due to
a unit load at the corresponding point of the beam. The line
joining the extremities of these ordinates is the "influence line"
for bending moments at F±.

I x
At a distance x from A2, the ordinate has the value ~, which

varies from o at A2 to -~ at F2. At a distance x from Z?2, the

value of the ordinate is -y-, which varies from o at B^ to -~

/ /

at F2. The influence line therefore consists of the two straight
portions A2FBt F3B2, where F2FB=-^. The ordinates are laid

off downward, the bending moment being always negative
according to the convention already adopted (Art. 47 and
Art. 1 1 6).

The influence line having been drawn, the bending moment
at F1 due to a load at any point J^ may be found by multiply-
ing the load by the corresponding ordinate J2JB. The bending
moment at Fl due to the combined action of several loads is
found by multiplying each load by the corresponding ordinate
of the influence line and adding the products.

Since the sign of the bending moment due to a load is the
same, whatever its position, and since the bending moment due
to a given load increases as the load approaches the section, the
following general principle may be stated :

The bending moment at any section has its greatest value
when the beam is as fully loaded as possible, with the heaviest
loads near the section.

This principle serves only as a rough general guide. An
exact rule will be deduced in the following Articles.

130. Load Polygon. — Let a curve or broken line be drawn,
of which the abscissa is horizontal, and the ordinate at any
point represents (on the assumed force scale) the total load up

BEAM SUSTAINING MOVING LOADS.

109

to that point, measured from some fixed point in the load series
Such a curve or broken line may be called a load polygon. In
the figure on PL IV, Q'R' is a load polygon for the series of
train loads shown on PL III, QR being a funicular polygon
for the same series.

A simple relation exists between the load polygon and the
funicular polygon, which is of use in the discussion of the prob-
lem now before us.

Let x and y be the coordinates of any point of the funicular
polygon, the jj/-axis (L Y) being vertical, and the -r-axis any hori-
zontal line LX. Let x1, y1 be the coordinates of the correspond-
ing point of the load polygon, referred to a vertical axis (L! Y'\
and a horizontal axis (L'X'} passing through the pole of the
force polygon used in the construction of the funicular polygon.

Consider any point of the funicular polygon, for example a

point on the string 02'. The value of -^ at this point is the

ctx

tangent of the angle between 02' and the horizontal. But since
02' is parallel to the corresponding ray of the force polygon,

this tangent is equal to ^, where H is the pole distance used

in drawing the funicular polygon. That is,

^=y

dx H'

131. Variation of Bending Moment at Any Section of a Beam.

— Let AlBl (at (A), PL IV) represent one position of a beam
of span /, and let Fl be any section at which it is desired to
study the variation of the bending moment. Let !1 = A1Fl,
/2 = F1B1, and z — horizontal distance of A1 from the vertical
axis L' Y1. Through Alt B^ and F^ draw vertical lines, inter-
secting the funicular polygon in A, B, F, and the load polygon
in A1, B\ F'. Let the vertical through F1 intersect AB in G,
and A'B' in G'. Let the ordinates of A, B, F, G, measured

1 10 GRAPHIC STATICS.

from LX, be denoted by a, b, f, g\ and the ordinates of A', Bf,
F', G', measured from L'X', by a' t £',/, g' .

If M denotes the bending moment at Flt and H the pole
distance,

-f); ......... (2)

and it will now be shown that the rate of change of this bend-
ing moment (i.e. the change per unit of horizontal displacement)
as the beam moves horizontally while the loads remain sta-
tionary is equal to F' G' or g' —f.

Since AGB is a straight line, and ^— =-i, it follows that

(3)
In an exactly similar manner it may be shown that

F>G'=g>-f'Jja'+fj6>-f ........ (4)

From (2) and (3),

(5)

If the beam moves horizontally, the abscissas of Av B^ and F1
change at the same rate. Differentiating (5) with respect to z,

_rr(^da Ldb__df\
\ldz ^ Idz dz)

da db , df „ dv

Now — , — , and -~ are equal to the values of ~ at A, B,
dz dz dz dx

and F, respectively; therefore, from equation (i) of Art. 130,
dz dz dz*

BEAM SUSTAINING MOVING LOADS.
dM L ,

hence

or, comparing with equation (4),

132. Condition for Maximum or Minimum Bending Moment at
Any Section. — As the beam moves relatively to the loads, FG
(proportional to M) in general varies. As it passes through a

maximum or a minimum value, must equal zero ; but, from

the relation just proved, this makes F'G' = o. That is,

When the bending moment at any given section of the beam is
a maximum or a minimum, the line A' B' intersects the load
polygon at a point directly opposite tJie section considered.

This condition may be stated in algebraic form as follows :
Let Wl denote the total load between A1 and Fv W2 the

load between F1 and B^ and W the total load on the beam.

Evidently,

If the points F1 and G' coincide, f — a' and b' —f are propor-
tional to A' G' and G'B* ', that is to /: and /2, and therefore,

/i ~ l

That is, when the bending moment at any section of the beam
is a maximum or a minimum, the average load per unit length
on each segment is equal to the average load per unit length on
the whole span.*

* It is to be noted that if, at any point, there is a load which is concentrated in the strict

mathematical sense (i.e., a finite load applied at a mathematical point), the value of —

dx
suffers a sudden change of value at the point of application of that load, and the equation

udy

y = H ,-

does not hold at that point. This consideration does not, however, invalidate the above
conclusions. For (a) a concentrated load in the strict mathematical sense does not occur

112 GRAPHIC STATICS.

133. Discrimination between Maxima and Minima. — The

magnitude of the bending moment is

it is obvious from the form of the funicular polygon that g is
always greater than f.

If g1 — f is positive, g— /(and therefore M) increases as z in-
creases (i.e., as the beam is displaced in the positive direction).
If g1 —f is negative, M decreases as z increases. Therefore,
as M passes through a maximum value, the sign of g1 — fr
changes from plus to minus ; and as M passes through a mini-
mum value, g' —f changes from minus to plus (the displace-
ment of the beam being in the positive direction, that is, toward
the right as the figure is drawn on PL IV).

If the load series consists of concentrated loads, g' —f will in
general pass through zero only when a load is at one of the
three points Alt Bv F^ By applying the test just given, it may
be seen that a maximum value can result only when a load
passes F±.

In applying the criterion for a position of maximum or mini-
mum value of the bending moment, and in distinguishing posi-
tions causing maximum stress from those causing minimum
values, it is not necessary to draw the line A' B' for every trial
position of the beam. A movable strip of paper representing
the beam may be laid in any desired position, the points A' and Br
approximately located by inspection, and a thread stretched
between them. If the thread crosses the load polygon at a
point directly opposite the section in question, the condition for

in practice, and ($) the ideal case of a finite load applied at a mathematical point may be
treated as the limit of a case of distributed loading, so that even in that case the conclu-
sions above derived hold, with slight change in the form of statement.

The series of concentrated loads actually coining upon a bridge in practice gives a load
polygon which, while approaching the form shown on PI. IV, differs from it in that the
portions under the loads are not strictly vertical lines, but are lines of variable (very steep)
inclination.

In applying the condition just deduced, a load at either of the three points A±, B^} F^t
is to be regarded as distributed in an arbitrary way along the small element of horizontal
distance near the line used as its line of action.

BEAM SUSTAINING MOVING LOADS. 113

maximum or minimum is fulfilled. Only positions which bring
a load at the section need be tried. The accurate location of
Ar and B' is usually unnecessary.*

At (7?), PI. IV, is shown the beam AlBl in such a position
that F' G' = Q. It is seen that this position corresponds to a
maximum value of the bending moment at F1 ; for a displace-
ment of the beam toward the right (positive) makes g3 — f
negative, while a displacement toward the left makes g' — f
positive.

1 34. Determination of Bending Moment. — Having determined
the position of the load series causing greatest bending moment
at a given section, the value of the bending moment may be
computed by the method already explained in the case of fixed
loads. At (M\ PI. Ill, is represented the position of the beam
A^B^ which causes the greatest bending moment at the section
Flt distant 16 ft. from Av the span being 64 ft. That the con-
dition for a maximum is satisfied in this position may easily be
verified. The total load on the beam is 112000 Ibs., one-fourth
of which is 28000 Ibs. By regarding 5000 Ibs. of the load 3-4
as belonging to the segment A-^F^ the load on this segment is
28000 Ibs. It is seen also that the beam is fully loaded, and
that the heaviest loads are near the section. Moreover, the
condition for maximum is not satisfied for any other position
near the one chosen.

The closing string of the funicular polygon for all external
forces acting on the beam in the position (M) is AB. The
bending moment at F1 is the product of the ordinate FG by the
pole distance.

At (/?), PL III, is shown the diagram of maximum bending
moments for all sections of the beam. From AlBl are laid off
downward ordinates proportional to the maximum bending
moments. The length FG just determined is laid off from F1 ;
this must be multiplied by the pole distance (100000 Ibs. in the

* See paper by Professor Henry T. Eddy, Trans. Am. Soc. C. E., Vol. XXII,

GRAPHIC STATICS.

diagram). The complete curve is drawn, but the construction
for determining values of the maximum bending moment is
shown only for the section F^.

135. Moment Curve for Fixed Loads. — The greatest bending
moment due to moving loads must be combined with the bend-
ing moment due to fixed loads. If the fixed load is uniformly
distributed, as already assumed in the computation of shear,
it may be divided into parts, each assumed concentrated at
its center of gravity, and a funicular polygon drawn, using the
same pole distance employed in the force diagram for moving
loads. The ordinates of this funicular polygon may be laid off
upward from the line A^B^, and their ends joined to form a
continuous curve. The total ordinate from this curve to that
already drawn for moving loads represents the true greatest
bending moment at the corresponding section of the beam.
The curve is shown at (B), PI. Ill, but the construction is
omitted, since it involves no new principle.

It is to be remembered that the bending moment found for
any section is a possible value for the other section equally
distant from the center of the beam, since the train may be
headed in the opposite direction, and the same construction
made, viewing the beam from the other side. The same state-
ment holds as to shears.

136. Design of Beam sustaining Moving Loads. — In designing
a beam to sustain moving loads, the greatest shear and bending
moment that can come upon it for any position of the loads must
be known for every section. The methods above given are
sufficient for the determination of these quantities ; and the
problem of designing the beam will not be here further discussed.

CHAPTER VII. TRUSSES SUSTAINING MOVING

LOADS.

§ i. Bridge Loads.

137. General Statement. — The most important class of trusses
sustaining moving loads is that of bridge trusses. The two
main classes of bridges are highway bridges and railway bridges.
The forms of trusses most commonly used differ for the two
classes, as do also the amount and distribution of loads.

Before the design can be correctly made, the weights of the
trusses themselves must be known. Since these weights de-
pend upon the dimensions of the truss members, they cannot
be known with certainty until the design is completed. The
remarks made in Art. 82 regarding the design of roof trusses
are here applicable.

In the following articles we shall give data available for
preliminary estimates of truss weights.

Railway bridges of short span are frequently supported by
rolled or built beams. When the span is longer than 100 ft.
trusses should be used. (Cooper's " Specifications for Iron and
Steel Railroad Bridges.")

138. Loads on Highway Bridges. — (i) Permanent load. —
The permanent load sustained by a highway bridge truss in-
cludes the weight of the truss itself, of the lateral or "sway"
bracing, of the floor and the beams and stringers supporting it.
These weights are all subject to much variation, but, for pur-
poses of preliminary design, the following formula, taken from
Merriman's " Roofs and Bridges," may be used.

H6 GRAPHIC STATICS.

Let w — weight of bridge in pounds per linear foot; /=span
in feet ; b = width in feet. Then

w= 140+ 12 b + o.2 £/— 0.4 I.

(2) Snow load. — The weight of snow may be taken as in
case of roof trusses (Art. 84). The values there given are
probably in excess of those ordinarily employed in practice.

(3) Wind load. — The pressure of wind striking the bridge
laterally is resisted by the chord members together with the
lateral bracing. These constitute horizontal trusses, in which
the stresses are to be found in the same way as for the main
trusses of the bridge. As the determination of wind stresses
requires the use of no special methods or principles, they will
not be here considered. The student is referred to Burr's
" Stresses in Roofs and Bridges," Merriman's " Roofs and
Bridges," and other available works for a complete discussion
of wind pressure and its effects on bridge trusses.

(4) Moving load. — The most dangerous moving load for a
highway bridge is usually a crowd of people or a drove of
animals. This is commonly taken as a uniformly distributed
load, which may cover the whole bridge or any portion of it.
Its value is variously taken at from 60 Ibs. to 100 Ibs. per square
foot of area of floor, depending upon the span and upon local
conditions.

It may be that in certain cases the greatest stresses will
result from the passage of heavy pieces of machinery over the
bridge, as, for example, a steam road roller. This should of
course be considered in the design.

For a complete discussion of loads on highway bridges, the
student is referred to Waddell's " Highway Bridges."

139. Loads on Railway Bridges. — (i) Permanent load. —
The permanent load on a railway bridge includes (a) the wei'ght
of the track system, which is known or may be determined at
the outset ; (b) the weight of longitudinal stringers and cross-

BRIDGE LOADS. 117

beams, which can be determined before the trusses are designed ;
and (c) the weights of trusses. The weight of the track system
may be taken at 400 Ibs. per linear foot for a single track.
(See Burr's " Stresses in Bridge and Roof Trusses " ; Cooper's
" Specifications for Iron and Steel Railroad Bridges " ; Merri-
man's " Roofs and Bridges.") The total weight of track system
and supporting beams and stringers varies from 450 Ibs. to
600 Ibs. per linear foot. (Merriman.) For spans less than 100
feet, Merriman gives the following formulas, in which w is the
total dead load of the bridge in pounds per linear foot, and / is
the span in feet :

For single track, w= 560+ 5.6 /.

For double track, w= 1070+ 10.7 /.
See also Art. 8 of Burr's work above cited.

(2) Snow and wind. — Railway bridges usually offer little
opportunity for the accumulation of snow. Wind pressure is,
however, an important factor. Besides the pressure upon the
bridge itself, the pressure upon trains crossing the bridge must
be considered. The latter is a moving load and may be dealt
with in the same way as other moving loads. Its amount may
be computed from the area of the exposed surface of the train
and the known (or assumed) greatest pressure due to wind
striking a vertical surface (Art. 85).

For further discussion of wind pressure, the student is referred
to the works already cited. The graphic methods of deter-
mining stresses due to wind will be evident when the methods
for vertical loads given in the following articles are understood.

(3) Moving loads. — The moving load to be supported by a
bridge consists of the weights of trains. Such a load is applied
to the track at a series of points, namely, the points of contact
of the wheels. But the load is applied to the trusses only at
the points at which the floor beams are supported. Hence the
actual distribution of loads upon the truss is somewhat com-
plex. It was formerly common to substitute for the actual

H8 GRAPHIC STATICS.

load a uniformly distributed load, thus simplifying the problem
of determining stresses. It is now more usual to consider the
actual distribution of loads for some standard type of locomo-
tive used by the railroad concerned, or specified by its engi-
neers. For examples of such distributions the student is
referred to Cooper's " Specifications " already cited ; also to
PL III, and to the following portions of this chapter.

140. Through and Deck Bridges. — A bridge is called through
or deck according as the floor system is supported at points of
the lower or of the upper chord. In the former case, if the
trusses are too low to require 'lateral bracing above, they are
called pony trusses.

The weight of the truss itself is to be divided between the
upper and lower joints. But the weight of the floor system
and of the supporting beams and stringers comes wholly at the
lower joints of a through bridge, or at the upper joints of a
deck bridge. The moving load is, of course, applied at the
same joints at which the floor system is supported.

If the floor system is supported directly upon the upper
chord, as is sometimes the case, the moving load and part of
the dead load produce bending in the chord members ; the
design is otherwise unaffected by this construction.

§ 2. Truss regarded as a Beam.

141. Classification of Trusses. — Since a bridge truss acts as
a practically rigid body resting on supports at the ends or other
points and sustaining vertical loads, it may be regarded as a
beam, and trusses may be classified in the same way as beams
(Art. 1 14). The only class to be here considered is that of
simple trusses, or such as may be regarded as rigid bodies for
the purpose of determining the reactions.

Cantilever trusses and continuous trusses are defined like the
corresponding classes of beams (Art. 114). The most impor-
tant case is that of a truss simply supported at the ends.

TRUSS REGARDED AS A BEAM.

IIQ

b i

Fig. 47

142. External Shear for a Truss. — • If a truss be regarded as
a beam, the external shear, resisting shear, and internal shear-
ing stress at any section may be defined just as in Art. 115.
In some forms of truss a knowledge of the external shear at
any section makes it possible to compute readily the stresses in
certain truss members. Thus, in the portion of a truss repre-
sented in Fig. 47, let the section
MN cut three members, of which
two are horizontal. (The member
x'y' is disregarded.) Since each
member can exert forces only in
the direction of its length, the ex-
ternal shear in the section MN
must be wholly resisted by the di-
agonal xy ; and the internal force
in xy must be such that its resolved

part in the vertical direction is equal in magnitude to the exter-
nal shear in the section. Let the external shear V be repre-
sented by YZ (Fig. 47) ; draw YX parallel to yx and ZX
horizontal; then XYwill represent the stress in xy. If V is
positive (Art. 115), the stress in xy is a tension. If Vis nega-
tive, the stress is a compression. If the member xy were
replaced by one sloping the other way from the vertical, these
statements as to kind of stress would be reversed.

If no two of the three members cut by any section are par-
allel, the stresses cannot be computed so simply, since all may
contribute components of force to resist the external shear.

143. Bending Moment for a Truss. — In many cases the
stresses in the truss members can be found from the values of
the bending moment at different sections.

Thus, in Fig. 49, let a section MN be taken cutting three
members as shown, and let the origin of moments be taken at
the point of intersection of two of them (as bx and xy) ; then
since the moments of the internal forces in these two will be

120 GRAPHIC STATICS.

zero, the resisting moment is equal to the moment of the inter-
nal force in the third member ay. The arm of this latter force
is the perpendicular distance of its action line from the origin.
Call it k, and let / represent the force itself, and M the bending
moment. Then, numerically,

ph = M. .'./=^-

If M is positive (in accordance with the convention of Art.
47), / must act from right to left, hence the stress in ay is a
compression. If M is negative, the stress is a tension. (It
must be remembered that the forces whose moments make up
the bending moment M act upon the portion of the truss to
the left of the section.)

§ 3. Truss sustaining Any Series of Moving Loads.

144. General Method of Determining Maximum Stresses in
Truss Members. — When a truss carries a definite series of mov-
ing loads, the problem of determining the maximum stress in
any member involves (i) the determination of the proper posi-
tion of the loads and (2) the actual computation of the stress for
a known position of the loads. The second part of this problem
involves only the principles already explained in the discussion
of trusses carrying fixed loads. The first part requires special
consideration.

Although the treatment of trusses with parallel chords is in
some respects simpler than that of the case of non-parallel
chords, it is advantageous to consider the problem in as general
a manner as possible. The following discussion will therefore
refer to the general case of a truss of any form, subject only to
the restriction that there are no redundant members, so that the
truss may be completely divided by cutting three members, of
which any chosen member may be one.* No restriction will be

* A case not included in this general description, while still offering a determinate
problem, is that of a truss with subordinate bracing. This important case will be consid-
ered in a following discussion.

TRUSS SUSTAINING ANY SERIES OF MOVING LOADS. 121

made as to the distribution of the loads. Trusses with parallel
chords, and particular distributions of loading, may be treated
as special cases of the general problem.*

145. Computation of Stress when Position of Loads is
Known. — The general " method of sections " (Art. 69) may be
applied to determine the stress in any member when the posi-
tion of the load series is known.

Thus, referring to PI. V, let 1-2, 2-3, . . . represent the lines
of action of the successive loads, 1-2-3- • • • the force polygon
for the loads, and let a funicular polygon be drawn as shown.
The truss is represented in two positions marked (A) and (B)
respectively. In connection with (A) is shown the construction
for determining the stress in the member C-JD^ and in connec-
tion with (B) the construction for determining the stress in C^H.
Referring to the former case, let a section be taken through the
three members C^D^ C^F^ H\F<i\ then the internal forces in
the three members cut are in equilibrium with the external
forces acting on the portion of the truss on either side (as the
left) of the section. These forces are the reaction at A^ and all
the loads from A± to C±. Applying the principle of moments, the
origin may be taken at F2, the point of intersection of the lines
C^F^ and H^F^ thus eliminating the forces acting in these lines.

The moment of any load is equal to the sum of the moments
of any components which may replace it. Hence for any load
between A± and C^ (though actually coming upon the truss at
the panel joints) the moment may be computed as if it were
applied to the truss at a point in the vertical line through its
actual point of application on the floor or track. But in the
case of a load on the panel C-J)^ only one component comes
upon the portion of the truss to which the moment equation is
to be applied ; it is therefore necessary to determine the portion
of every such load which is actually carried at Cv If loads
between 6\ and D1 are carried wholly at the two points 6\ and

* The case of parallel chords, with graphic methods of treatment, receives very full dis-
cussion in a paper by Professor Henry T. Eddy, Trans. Am. Soc. C. E., Vol. XXII (1890).

122 GRAPHIC STATICS.

D^ the portion carried at each of these points is easily deter-
mined, as will be shown.*

From Al and B^ draw vertical lines intersecting the funicular
polygon at A and B respectively ; the line AB is the closing
line of the funicular polygon for the external forces acting upon
the truss when the load series has the assumed position. On
the panel C^D^ is the load 6'f. Treating C1D1 as a simple
beam, the components replacing this load at Cl and D1 are found
(Art. 45) as follows: Draw vertical lines through C1 and Dl
intersecting the strings 06' and of in C and D ; from the pole O
draw the ray OK parallel to CD ; the required components of
6'f are represented in the force polygon by & K and Kf . The
same construction might be made for every panel, and it is seen
that the series of lines such as CD would form the funicular
polygon for the loads as actually applied to the truss at the
several joints. But the construction is here needed only for
the panel C^D-^ since a load on any other panel comes wholly
upon one of the two portions into which the truss is divided by
the section taken.

The sum of the moments of the external forces acting upon
the left portion of the truss may be found graphically by the
method of Art. 56. That is, through the origin of moments a
line is drawn parallel to the resultant of the forces (a vertical
line in this case) ; its intercept FG, between the strings AB and
CD, multiplied by the pole distance, gives the required moment.
The origin being taken at F^ this moment is equal to the
moment of the internal force in the member C1Dl ; hence that
force may be found by dividing the moment by the perpendicu-
lar distance from F2 to C-JDV

Referring next to (B) and considering the member C-^H, the
stress in this member may be found in a similar manner. The

* It is to be noticed that the actual distribution of any load among the different joints is
not determinate by simple methods, if the longitudinal beams supporting the floor system
are supported at more than two points. It will here be assumed that the portion of such a
beam between two supports acts as a simple beam. The results of this assumption are
probably as reliable as could be obtained by a more elaborate discussion.

TRUSS SUSTAINING ANY SERIES OF MOVING LOADS. 123

points A, B, C, D, F, G have the same meanings in this con-
struction as in the preceding, but AB and CD must be produced
in order to find the intercept FG, because the origin of moments
is at Flt beyond the end of the span. The moment of the
internal force in C^H is equal to FG multiplied by the pole dis-
tance; dividing this moment by the perpendicular distance
from F1 to C^H gives the magnitude of the required force.

The kind of stress in any member may be determined by
inspection of the conditions, as in Art. in.

If the stress in a member is determined by the foregoing
method, the steps of the process are the same, whatever mem-
ber be under consideration. These steps are as follows :

(1) A section is taken completely dividing the truss and cut-
ting only three members, one of which is the member whose
stress is to be determined, and one a member of the chord car-
rying moving loads.

(2) Taking origin at the intersection of those two of the
three members cut whose stresses are not in consideration,
equate the sum of the moments of the external forces acting
upon one portion of the truss to the moment of the internal
force in the member considered.

146. Definitions and Notation. — The following definitions
and notation will conduce to clearness and generality in the ensu-
ing discussion.

Let the origin of moments, assumed in determining the stress
in any member by the foregoing general method, be called the
moment-center for that member.

Let the stress-moment for any member be defined as the mo-
ment of the stress in that member with respect to the moment-
center just described. Let the sign of the stress-moment be
specified in the following manner :

Rotation being called positive or negative in accordance with
the convention adopted in Art. 47, let the stress-moment be
called positive when it resists a tendency of the left portion of

I24

GRAPHIC STATICS.

the truss to rotate negatively (and of the right portion to rotate
positively).

In all cases let the left and right ends of the span be marked
Al and Blt and let 6\ and Dl designate the left and right ends
of the panel cut by the assumed section. Let the vertical line
through the moment-center cut A1Bl (produced if necessary) in
a point marked Fv

Distances from left to right being positive, and from right to
left negative, let the following notation be used :

\ = n-

The sign of each of the four quantities /lf /2, «lf n% is to conform
in all cases to the direction, as indicated by the order of the letters
used in denning it. Thus !1 = A1F1 is positive or negative,
according as F1 is to the right or to the left of Ar

147. Effect of a Single Load. — Influence line. — A general
idea of the position of a load series which will cause the greatest
stress in any member of a truss may be obtained by considering
the effect of a single load placed in any position. The details
of the discussion will vary with the form of the truss, but the
method of reasoning may be sufficiently illustrated by reference
to the form shown at (A) or (B), PI. V.

The relative effect of loads in different positions in producing
stress in a given member may be clearly represented by means
of an " influence diagram," analogous to the diagrams used in
the discussion of shear and bending moment in a beam (Arts.
126 and 129).

From a given base line let ordinates be erected such that the
ordinate at any point represents (to a convenient scale) the stress
due to a unit load at the corresponding point of the span. The
line joining the extremities of these ordinates is the influence
line * for the member in question.

* For a discussion of influence lines, see paper by Professor George F. Swain, Trans.
Am. Soc. C. E., Vol. XVII, p. 21 (1887).

TRUSS SUSTAINING ANY SERIES OF MOVING LOADS. 125

It will be shown that there is a simple method, of complete
generality, by which the influence line may be drawn for any
member whatever. In order to explain this method clearly, it
is necessary to adopt a convention as to signs.

In laying off ordinates of the influence line, let positive values
be drawn upward and negative values downward from the refer-
ence line. Let the sign of the ordinate agree in all cases with
that of the stress-moment (Art. 146). With this convention, a
positive ordinate may correspond to either a compressive or a
tensile stress, depending upon the relation of the moment-center
to the member under consideration. From the sign of the stress-
moment the kind of stress may always be readily determined by
inspection of the truss diagram.

In Fig. 48 (PI. VIII) are shown the forms assumed by the
influence line for three typical members of the truss shown ;
(A) represents the case of the chord member C^D^ (B} that of
the chord member C^D^ and (C) that of the web member
C^DV The method of constructing these diagrams will now
be explained.

Member of loaded chord. — Consider first the member C^D^
The moment-center is C^, and Fl falls between C± and Dlf the
corresponding position of Fz being shown at (A). Here /j, /2,
nv n2 are all positive (Art. 146). A load P between D± and Blf

distant x from Bv causes a reaction — — at Alt and a stress-

Pl x

moment equal to — J— . This varies directly as x, being o at Bl

and — *( 2-^2) at 2}lm Similarly, the stress-moment due to a

PI x
load P between A1 and Cv distant x from Alf is — £-, which

varies from o at Al to — 2V i~^i) at ^ jf the moment-arm

of the required stress is t, the ordinates of the influence line at
C and D are

and the straight lines A2CS and D^B^ are the portions of the

126 GRAPHIC STATICS.

influence line for the corresponding portions of the span. A

load P between C^ and Dv distant x from Clt comes upon the

p (n _ x\
truss partly at 6\ and partly at Dl ; the former part is — ^ -- 4

Px n

the latter — . The total stress-moment due to such a load is

therefore

P(n-x) /a(/i

n In

putting P=\ and dividing by t, this expression gives the ordi-
nate of the influence line for any point between C-^ and Dv The
influence line for this portion of the span is therefore the straight
line C^DB. It will be noticed that, if the straight line B2D3 be

continued, its ordinate at F2 will be F2f2 = -~\ if A2C3 be con-
tinued, its ordinate at F2 will have the same value.

Member of chord not carrying moving loads. — Next consider
the member C^D^. The moment-center is Dv Fl falls at Dv
and F2 coincides with D2 as shown at (B\ Fig. 48. If the
reasoning used in the preceding case be repeated, it is seen that
the values there given for the ordinates of the influence line
apply also to this case. Since in the present case n2 = o, the
value of the ordinate at D2 reduces to -^-1. Further, it is seen
that A2CSD3 is a straight line.

Web member. — For the member Cl'Dl the moment-center is
at the intersection of C^D± and C1D1 ; F± falls to the left of Alt
the corresponding position of F2 being shown at (C), Fig. 48.
Here it will be noticed that /x and n^ are negative. If, however,
the method used in the preceding cases is again applied, it is
found that the values above given for the stress-moment due to
loads on D^B^ and A^C^ are correct for the present case both in

magnitude and in sign. Thus, for a load P on D^B^ distant

x from Blt the reaction at A^ is — , and the stress-moment is

PI x

— L. This stress-moment is negative (according to the conven-
tion adopted in Art. 146) ; but since /x is negative, the expres-

TRUSS SUSTAINING ANY SERIES OF MOVING LOADS. I2/

PI x

sion — L_ is correct in magnitude and sign. For a load P on

^(Tp distant x from ^ the stress-moment is positive, and is

given in magnitude and sign by the same expression as in the

PI x

preceding cases, — %—. The ordinates at C2 and D2 also have
/

the general values

the latter expression being negative (because /: is negative and
/2 — ^2 positive) while the former is positive (because /2 and l^ — n^
are both positive). Further, if A2C3 and D^B^ be prolonged,
they have a common ordinate -L* at the point F2. This ordi-
nate is negative, agreeing in sign with /r

General result. — It is thus seen that the construction of the
influence line may follow one general rule in all cases. This
rule may be stated as follows, designating points by letters, as
above :

From the point F2 erect an ordinate of magnitude and sign

-yA From the extremity of this ordinate draw straight lines to

A2 and #2, the former intersecting the vertical through C2 in Cs,
the latter intersecting the vertical through D2 in D3. The line
A^C^D^B^ is the required influence line.*

148. Approximate Position of Load-Series causing Greatest
Stress in a Given Member. — Inspection of the influence dia-
gram for a given member enables the position of loads causing
maximum stress to be estimated approximately.

Member of chord carrying moving loads. — For the member
ClDl (Fig. 48, PL VIII), the influence diagram (A) shows that

* The same general rule applies when FI falls between A± and C\, or between D^ and
/?!, «! being negative in one case, and »2 *n ^e other, while both /x and /2 are positive in
both cases. These cases occur in no ordinary form of truss, except that with subordinate
bracing. This form is discussed in Art. 160, and the influence diagrams for negative val-
ues of #! and of n.2 are shown on PL VIII.

128 GRAPHIC STATICS.

all loads on the truss produce the same kind of stress, and that
the effect of a given load is greater the nearer it is * to Dlf
Therefore, with any given load series, it is to be expected that
the greatest stress will occur when (a) the truss is heavily
loaded throughout the span, and ($) the heaviest loads are near
the point D^.

Chord not carrying moving loads. — In a similar manner, the
influence diagram (B) shows that the greatest stress in C^D^ is
to be expected when (a) the entire span is heavily loaded, and
(b] the heaviest loads are near Dr

Web member. —Diagram (C\ Fig. 48, PL VIII, shows that
for loads between A1 and £l the stress-moment is positive,
while for loads between E^ and B^ it is negative. Hence for
the maximum stress of the kind corresponding to a positive
stress-moment (a) the portion AlEl should be heavily loaded
while little or no load is between £1 and Bl9 and (b) heavy loads
should be near Cv For maximum stress of the kind corre-
sponding to a negative stress-moment, (a) E^B-^ should be
heavily loaded while little or no load is on A-^E^ and (b) heavy
loads should be f near Z>r

The above principles serve as an approximate general guide,
but do not enable us to determine the position of loads for
maximum stress exactly. A more definite discussion of the
problem will now be given.

149. Criterion for Position of Loads causing Maximum or
Minimum Stress in Any Member of a Truss. — Let a funicular
polygon and a load polygon be drawn for the given series of
loads (PL V).

As shown in Art. 130, if y denotes any ordinate of the
funicular polygon, and y1 the corresponding ordinate of the

* It may, of course, happen that C2Q is greater than D2DB. In such case, the effect
of a load is greater the nearer it is to Q.

f The nature of the load-series may be such that a compromise must be made between
requirements (a) and (£).

TRUSS SUSTAINING ANY SERIES OF MOVING LOADS. 129

load polygon (the axes being LX, LY, and L'X', LfY!,

respectively),

Consider the stress in any member of the truss shown at (A),
PI. V, as C-J}^ Taking a section through C^D^ C\F<& an<^ ^i^2»
the origin of moments for computing the required stress (the
moment-center) may be taken at F2. Let vertical lines through
Av Bv Cv Dl intersect the funicular polygon in A, B, C, D,
and the load polygon in A', B1, C', D' . Through Fz draw a
vertical line intersecting AB in G and CD in F; then (Art. 145)
the moment of the required stress with respect to the origin F2.

where H is the pole distance used in drawing the funicular
polygon. If the load series moves (or if the truss is considered
as moving while the loads remain stationary), the stress in C1D1
is always proportional to FG, and when the stress is a maxi-
mum or a minimum, the intercept FG is also a maximum or a
minimum.

Let the vertical through Fl intersect A'B' in G' and C'D' in
Ff. It will now be shown that, as the truss moves horizontally,
the change of FG (and therefore of M) per unit distance moved
is always proportional to F'Gf. The reasoning is similar to
that employed in Art. 131.

Let a, b, c, d, f, g, denote the ordinates of A, B, C, D, F, G,.
respectively, measured from the horizontal axis LX\ and a\ b' ,
cf> d',f\ g\ the ordinates of A, B', C, Z>', Ff, G', measured
from the axis L' X' . Also, let z denote the abscissa of the point
Alt measured from the axis L' Y!, and let /1? /2, /, nlt n^ n have
the meanings assigned in Art. 146.

It may be shown as in Art. 131 that

I3o GRAPHIC STATICS,

and by exactly similar reasoning,

n n
Therefore

= H(g-f)=H(a + llb-^c-^d. . (2)

Similar reasoning applied to the lines A'B', CD' leads to the
equation

The rate of change of Mas the truss moves horizontally is found
by differentiating (2) with respect to z. Observing (as in Art. 131)

that -A — , ... are the values of -/ at A, B, . . ., equation (i)
dz dz dx

shows that

rjda_ f Tjdb _ if

dz dz

the differentiation of equation (2) therefore gives

__.

dz I I n n
Comparing with (3),

^=g'-f' = F'G< (4)

It follows from equation (4) that when the position of the
loads is such that M is a maximum or a minimum (that is, when
the stress in ClDl is a maximum or a minimum), the ordinate
F'G' is equal to zero.

The above reasoning is general,* and the conclusion applies
to any member of the truss. In the case of chord members the
point Fl will in general fall within the span ; only in exceptional

* That the above reasoning is rigorous for the case of concentrated loads as actually
applied to the truss in practice follows from the considerations mentioned in the note in
Art. 132. With slight change in the form of statement, the conclusion holds even in the
ideal case of concentrated loads in the strict mathematical sense.

TRUSS SUSTAINING ANY SERIES OF MOVING LOADS. 131

forms of truss will it fall without the panel C-J)^ though it may
coincide with C^ or Dv The figure for the case of a web mem-
ber is shown at (B\ PL V, the letters A, B, C, £>, F, G, with or
without subscripts or accents, denoting corresponding points in
the constructions shown at (A) and at (B). The point Fl gener-
ally falls without the span in case of a web member; it is possi-
ble that it should fall between Al and C± or between D^ and B±.
In any case equation (4) holds, and the condition for maximum
or minimum stress takes the above general form.

The criterion above reached (F'Gr = o) is easy of application.
Let a strip of cardboard be prepared, showing the truss-skele-
ton and the moment-centers for all members of the truss. By
applying this to the diagram showing the load polygon, the
points A' and B' may be located approximately by inspection
(an accurate determination of them not usually being necessary
for the purpose in hand); by stretching a thread through A'
and B1 , the point Gr may be located approximately ; then, hold-
ing one point of the thread at G1, it may be stretched so as to
pass through one of the points C' and Df, and if it passes
through the other also, the condition for maximum or minimum
stress is satisfied. As will be shown presently, when the load-
series consists of concentrated loads, a load will generally be at
C± or Dl when the stress is a maximum ; the load polygon will
therefore be vertical (or very steep) at C' or at D't so that one
of these points may be shifted vertically without changing
appreciably the position of the other or of either Ar or B' . In
some cases (but not usually) it may be necessary to locate A',
£', and Gr accurately.

150. Positions giving Maximum and Minimum Values of the
Stress Distinguished. — If g—f and gl — f have like signs, the
magnitude of g—f increases with a positive displacement, and
decreases with a negative displacement; if they have unlike
signs, g—f decreases with a positive displacement, and increases
with a negative displacement. It follows that, if the truss is in

I32 GRAPHIC STATICS.

a position of maximum stress for the member considered, a
small positive displacement makes the signs of g—f and g' — f
unlike, while a small negative displacement makes their signs
like ; while the reverse is true for small displacements from a
position of minimum stress. (The words " maximum " and
" minimum " here refer to magnitude, irrespective of the nature
of the stress.)

A complete discussion of this principle would require the
consideration of various forms of truss. It will be sufficient
to illustrate the application to the form shown on Plates V
and VI.

151. Application of General Principles. — On PI. VI is repre-
sented a truss of 60 ft. span, divided into six equal panels.
The greatest depth of the truss is 12 ft.; the lower chord
(carrying the moving loads at the joints) is straight, while the
upper joints lie upon the arc of a circle. The two web mem-
bers intersecting at any upper joint are of equal length and
slope.

The moving load consists of the two locomotives and train
represented on PI. Ill, for which a load polygon and a funicular
polygon are drawn on PI. VI. The diagram represents half the
actual weight of the train, it being assumed that the load is
divided equally between two trusses.

The construction for determining the greatest stress in the
chord member C1D1 is shown at (A), PI. VI, and the construc-
tion for the case of the web member HDl is shown at (B). Like
lettering is used in the two cases.

Member of lower chord. — From the approximate general rule
deduced in Art. 148, it follows that, in order that the stress in
ClDl may have its greatest value, the truss must be loaded as
completely as possible. Moreover, since the effect of a given
load is greater the nearer it is to the panel C-J)^ the positions
first to be tested are those bringing the heaviest loads upon or
near the panel.

TRUSS SUSTAINING ANY SERIES OF MOVING LOADS.

133

The condition for maximum or minimum stress is found to
be satisfied in the position in which the truss is shown in the
figure, the heavy load 5 '6' being at D1 ; but not in any other posi-
tion in which the heaviest loads are upon or near the panel C1D1
and the truss fully loaded. The criterion of Art. 150 shows
conclusively that this position gives a maximum and not a mini-
mum value of the stress ; for g—f is positive, and a positive
displacement of the truss makes g1 —f negative, while a nega-
tive displacement makes it positive.*

The value of the maximum stress in C^D-^ may now be deter-
mined. The pole distance is 100000 Ibs. ; FG = q.g ft.; F(fl
(the moment-arm of the stress) is 12 ft. The required stress is
therefore a tension of

9.9xIOoooo=8 bg

Web member. — Referring to (B\ PL VI, consider the mem-
ber HD^. From Art. 148 it is known that when this member
sustains its greatest tension the portion DlBl must be as com-
pletely loaded as possible (with A^C^ probably free from loads),
while to produce the greatest compression A-f^ must be as
fully loaded as possible (with D1B1 probably free from loads).
Whether, in either case, loads should be between C^ and D± is
not known without applying an exact test.

Considering the case of tension, let the load series be brought
from the right until the foremost load reaches the panel, and
let the criterion for maximum stress be applied to successive
positions until one is found in which it is satisfied. It is found
that the criterion is satisfied when the second load (2-3) is at
Z\. That this position gives a maximum and not a minimum
value of the stress is found by applying the test of Art. 150.
For g—f is negative ; a positive displacement of the truss

* It will be noticed that, with a series of concentrated loads, F'G' will, as a rule, pass
through zero only when a load passes one of the four points Alt Blt Q, D±. Further,
when F' is between C and D' , a load at AI or B^ (when F'G'=o) will correspond to a
minimum stress, while a load at C\ or D± will correspond to a maximum.

134 GRAPHIC STATICS.

makes g'-f positive, while a negative displacement makes it
negative.*

The value of the maximum tension in the member HDl may
now be found by the method of moments. The intercept
FG = 8.8 ft.; the pole distance is 100000 Ibs. ; the arm Flm/
= 39.6 ft. The required tension is, therefore,!

8.8 x i ooooo

39-6

= 22200 Ibs.

152. Algebraic Statement of Condition for Maximum or Mini-
mum Stress. — The general condition above deduced admits of
simple algebraic expression. For this purpose let

PI = total load on A^C^
P2 = total load on D^B^
Q = total load on C-J)^
W= total load on AB

Let /j, /2, «1} «2, /, n have the meanings assigned in Art. 146,
their algebraic signs being carefully observed.

Referring to Fig. (A\ PI. V, the point G' divides A' B' into
segments proportional to ^ and /2, and Fr divides C'D' into seg-
ments proportional to n± and ;/2. Also, with the notation of
Art. 149, it is seen that

Since A'G'B' is a straight line,

g'-a1 _b!-?' _b'-a' _ W

/i ~~4~~~7~ "~r*

* Here again it is seen that the condition for maximum or minimum stress will, in
general, be satisfied only when a load is at one of the four points Alt Z?1( Q, D^. In this
case a maximum stress of the kind now under consideration will occur only when a load
is at £>!.

t These results, being obtained from small-scaled drawings, are not given as accurate
values of the stresses, the present object being merely to illustrate the method of pro-
cedure. By careful work, and with drawings made to a suitable scale, a sufficient degree
of accuracy may be obtained by the above method.

TRUSS SUSTAINING ANY SERIES OF MOVING LOADS. 135

, , Wl±
hence g'—a'=—j±, (i)

Similarly, since C'F'D1 is a straight line,

f^cf^'-f' = df-Cf=Qf

HI 722 n n
hence f — cf = — -; (3)

Subtracting equation (3) from equation (i), and (4) from (2),
and writing P1 and P2 instead of c' — a' and b' — d' , there result
the equations *

If the points F' and G' coincide, as is the case when the con-
dition for maximum or minimum stress is satisfied, g' — f = o,
and equations (5) and (6) may be written |

W x,

- 7 - = 7 =~~T ......... ^''

L\ /2

Although the foregoing discussion has referred directly to the

* Equations (5) and (6) are not independent ; either may be derived from the other by
means of the relations PI + P2 + Q = W, /x + /2 = /, HI -\-n2 = n.

f It is interesting to compare this result with the condition for maximum or minimum
bending moment for a beam, given in Art. 132. If the portion of the load polygon from
C to D' were replaced by the straight line CD' (i.e. if the load Q were uniformly dis-
tributed over the panel), the total load from A1 to F1 would be/J1 + J Q, and the total load
from FI to B^ would be /2 + — Q. The condition for greatest stress in C^Dl is therefore
identical with the condition for maximum bending moment at Flt on the assumption that
the load Q is uniformly distributed. This interpretation of equation (7) is of use only
when F1 falls between C± and Z>lt although it holds in a mathematical sense in all cases.

136 GRAPHIC STATICS.

case in which Fl falls between Cl and D^ so that flt /2, ;/lf n2 are
all positive, the result holds also when any of these quantities
are negative, signs being duly observed in substituting their
values in equation (7).

If the member considered belongs to that chord to which the
moving loads are applied (as at (A), PI. V), /x and L will be
positive, and one or both the quantities n± and ;z2 will be posi-
tive. If one of the members C^F^ F^D^ were vertical, either
n^ or ;/2 would be zero. In no truss of ordinary design, without
.subordinate bracing, would n± or n2 be negative in case of a
chord member.

In case of a member of the chord carrying no moving loads,
the point Fl falls at one end of the panel C^D^ and either n^ or
;/2 becomes zero. This may be seen by referring to the member
H&, PL V, (A).

In case of a web member (as C^H, PL V, (-#)), F1 falls without
the panel ClDl (making n± or «2 negative); in the form of truss
here figured Fl also falls without * the span A^B^ (making /x or /2
negative). In this case equation (7) may be put into more con-
venient form. Consider, for example, the maximum tension in
the member D^H, in the truss shown at (B\ PL VI. From the
general principles stated in Art. 148 it follows that loads on
-Di-Bi cause tension in this member, while loads on A^C^ cause
compression. If possible, therefore, equation (7) should be
.satisfied by a position making P1 zero or small. The equation
may be written

and this is to be satisfied by a position making P^ = o or as
small as possible.

Example. — Test whether the general equation (7) or (8) is
satisfied in the positions shown on PL VI.

* A design is possible which will cause Flt for certain web members, to fall within the
span but without the panel C^D^. This does not occur in any ordinary form of truss.

TRUSS WITH SUBORDINATE BRACING.

137

153. Truss with Parallel Chords. — The application of the
results of the foregoing discussion to the case of trusses with
parallel chords presents no difficulty.* The only matter calling
for special remark relates to the case of web members.

Referring to (B\ PL VI, it will be seen that if the chords
were parallel the point F1 (and therefore also F and Ff) would
fall at infinity. In order that C' D' and A'B' might intersect in
Ff, these two lines would therefore need to be parallel. The
condition for maximum stress in a web member therefore
reduces, in case of parallel chords, to the condition that A'Bf
and CD' shall be parallel.

Further, the intercept FG and the moment-arm F^J become
infinite if the chords are parallel, so that the moment-equation
takes an indeterminate form. The stress in the member HD^
can, however, be easily determined by resolving vertically all
forces acting upon the portion of the truss to the left of a
section cutting the member in question and two chord members.
See Art. 142.

The algebraic formula expressing the condition for maximum
stress in a web member (equation (8), Art. 152) reduces to
simpler form in case of parallel chords. If Fl passes to infinity,

/x and «! approach a ratio of equality while -- approaches zero ;

n\
so that the equation becomes, in the limit,

The load P1 does not enter this equation, but is to be taken as
small as possible in applying it.

§ 4. Truss with Subordinate Bracing.

154. Description of Truss. — In Fig. 49, PI. VIII, is repre-
sented a form of truss designed to offer points of support for
the floor system intermediate between the main panel joints.

* For an exhaustive treatment of this subject consult a paper by Professor Henry T.
Eddy, Trans. Am. Soc. C. E., Vol. XXII (1890).

138 GRAPHIC STATICS.

The effect of the subordinate vertical members a and diagonal
members b requires some explanation. With the arrangement
shown in Fig. 49, the main diagonals are tension members and
the main verticals compression members, while the subordinate
verticals sustain tension and the subordinate diagonals compres-
sion. In the form shown in Fig. 50 (PI. VIII) the subordinate
diagonals sustain tension. The members represented by the
broken lines in both Fig. 49 and Fig. 50 do not act when the
truss sustains dead loads only. Such members will in general
be needed only in panels near the middle of the span. In
Fig. 49 the two upper diagonals (and in Fig. 50 the two lower
diagonals) in each main panel are designed for tension only ;
the one represented by the broken line comes into action only
when the loading is such that the other would be thrown into
compression.

155. Determinateness of Stresses. — Disregarding counter-
braces, it is easily seen that the stresses in all members are
determinate.

In the case of members such as e, f, or g (Fig. 49), the deter-
minateness follows from the fact that any such member can be
made one of three through which a section may be passed com-
pletely dividing the truss.

The stress in g1 is determinate, being always equal to that in g.
For a like reason, the stress in a is determinate, being always
equal to the load carried at Cv

Considering the point of intersection of a, b, d, e, since the
stresses in a and e are determinate, it follows that the stresses
in b and d are also determinate ; the force polygon for any four
forces in equilibrium being completely determinate when two of
the forces are known completely, and the directions of the other
two are given. This shows that the stresses in b and d would
be determinate even if d and e were not collinear.

Like reasoning applied to an upper joint shows that the
stress in any main vertical h is determinate.

TRUSS WITH SUBORDINATE BRACING.

139

156. Effect of Subordinate Braces on Stresses in Main Mem-
bers. — Upper chord. — It may be shown that the stresses in the
upper chord members have, for any position of the loads, the
same values they would have if the subordinate members were
not present. Thus, applying the method of moments in the
usual manner, the center of moments for the member /* is
the point Dv The stress is determined by dividing the truss
by a section through/, <?, and g, and applying the conditions of
equilibrium to either portion of the truss. For a load between
A1 and F^ consider the right portion of the truss. The reaction
at B1 is the same, whatever the division into panels ; hence the
stress is the same as if the members a and b were not present.
For a load between B1 and Dlt the same conclusion is reached
by considering the left portion of the truss. For a load between
jpj and D^ the moment equation for the right portion of the
truss will contain the reaction at Blt and the portion of the load
carried at Dl ; and this portion is changed by the subdivision
of the panel F-^D^ The moment of the load at Dl is, however,
zero ; so that the subdivision of the panel has no effect upon
the moment equation nor upon the stress in the member/.

Web members. — At any upper joint of the truss four forces are
in equilibrium, — two due to the chord stresses and two to the
stresses in the web members. Any two of these being given,
the other two are determined. Since the chord stresses have
the same value whether the subordinate members are present
or not, the same is true of the web stresses. In determining
the stresses in d and //, the subordinate members a and b may
therefore be disregarded.

The stress in e is changed by the presence of the subordinate
members if any load is between F^ and D^. The value of the
stress, may, however, be determined by the method of sections,
the center of moments being at the intersection of / and g.
The construction shown at (B\ PI. VI, may be applied to the

* The present discussion refers always to Fig. 49, PI. VIII.

I40 GRAPHIC STATICS.

member <?, the short panel C1Dl (Fig. 49) corresponding to the
panel ClDl in PI. VI, and the point Fl being at the intersection
off and g.

Lower chord. — The analysis just applied to the member e is
applicable also to a lower chord member, such as g. Taking a
section through/, e, and^-, the origin of moments is to be taken
at the intersection of e and f. The construction shown on
PL VI may therefore be applied, the panel marked ClDl in
Fig. 49 corresponding to ClDl in PI. VI.

Since the load at C^ and the member a are collinear, the
stress in g1 is always equal to that in g.

157. Condition for Maximum Stress in Main Truss Members. —
From the results of Art. 156, it follows that the method already
explained for determining what position of the loads causes a
maximum stress may be applied to all the main members of the
truss shown in Fig. 49. The condition in all cases is that the
points Ff and G1 , determined as on PI. V or PI. VI, shall coin-
cide. In case of certain members, however, the position of the
line C'D' is changed by the presence of the subordinate mem-
bers. To summarize :

In the case of the members /, d, and h (Fig. 49), the subordi-
nate members are to be disregarded in applying the condition
for maximum stress ; the whole panel F^D^ takes the place of
ClDl in the construction on PL VI. This principle applies to
all upper chord members, all main verticals, and the upper por-
tions of all main diagonals.

In the case of the members e and g, the short panel marked
ClDl in Fig. 49 takes the place of ClDl in the construction on
PL VI ; this construction being otherwise unchanged. This
applies to all lower chord members, and to the lower portions
of the main diagonals.

There is no difficulty in applying similar reasoning to the
truss shown in Fig. 50-

TRUSS WITH SUBORDINATE BRACING. 141

158. Subordinate Members. — When the position of the load
series is known, the stresses in the subordinate members (a and
b, Fig. 49) are easily determined.

The stress in a is always equal to the load carried at Cv

The members d and e being collinear, the resolved parts
of the stresses in a and b perpendicular to d and e must be
equal ; the stress in b can therefore be found directly from that
in a.

It remains to determine the position of the load series which
makes the load at £\ (and therefore the stresses in a and b) a
maximum.

Assuming that a load on any panel is carried to the truss
Wholly at the ends of that panel, the condition for maximum
load at any joint may be deduced without difficulty.

In Fig. 51 are represented the load polygon and funicular
polygon for a series of loads, AlFl and F-^B^ being consecutive

Fig. 51

panels of any lengths. Vertical lines through Av B^ and F1
intersect the funicular polygon in A, B, F, and the load polygon
in A', B' , F'. The portion of the loads on A^F-^ which is car-

1 42 GRAPHIC STATICS.

ried at F1 is found by drawing the closing string AF, and
parallel to it the ray OK'; the portion carried at F± is Kr2.
Similarly, drawing OK" parallel to the closing string FB, 2 K"
represents the portion of the loads on F1B1 which is carried at
Fv The total load carried at F1 is therefore K'K".

Prolonging BF to L, AFL and K'OK" are similar triangles;
hence AL and K'K" vary in the same ratio, as the load series
moves. But if G is the point in which a vertical through F
intersects AB, it is seen that FG varies as AL. Therefore,
when K'K" is a maximum, FG is a maximum.

Reasoning as in Arts. 131 and 132, it is found that when FG
is a maximum or a minimum the points Ar, F', and B1 are
collinear.

The condition for greatest load at F1 is therefore exactly
analogous to the condition for greatest bending moment at F1
on the supposition that A^B^ is a simple beam supported at
Al and BY

From this analogy the conditions for greatest stresses in the
members a and b (Fig. 49) may be stated as follows : *

(a) The panels F1C1 and C-J)^ must be loaded as heavily as
possible, the heaviest loads being near Cv

(b) The loads on the panels must be proportional to their
lengths.

In satisfying requirement (b\ a load must generally be at Cv
and must be regarded as divided between the panels in some
arbitrary manner.

159. Effect of Counter braces. — The foregoing discussion and
conclusions need modification when the loading is such as to
bring counterbraces into action. It is needful to determine
what counters are needed, what maximum stresses they sustain,
and whether the maximum stresses in any other members are
changed by their action. Let it be assumed that when the

*See paper by Professor H. T. Eddy, Trans. Am. Soc. C. E., Vol. XXII.

TRUSS WITH SUBORDINATE BRACING.

143

train moves from right to left the members called into action
are those represented in Fig. 52. The conclusions drawn in
Art. 155 as to the determinateness of the stresses are equally
applicable to the present case ; but the fact that the mem-
bers d and e are not collinear makes it necessary to modify

Fig. 52

the methods of determining greatest stresses in certain
members.

In case of the members / gt e, and a (Fig. 52) the usual
methods obviously apply.

If b is designed for tension only, it is necessary to determine
whether it acts when the stress in d is greatest. If it does not,
the discussion of d is obvious. If b does act, the stress in d is
not the same as in the case in which d and e are collinear. It
may be shown, however, that the stress in d always bears a
constant ratio to that which would exist if d and e wrere collinear.
This becomes evident if the force polygon be drawn for the
forces due to the stresses in / /', d, and //, and the effect of
changing the direction of d be observed ; it being remembered
that such change does not affect the stresses in//'.

In a truss of long span, the stress in d will usually be small,
since the dead load will be so great as to nearly or quite coun-
terbalance the greatest live load effect. Counters will be needed,
if at all, only in panels near the middle of the span.

The determination of the maximum stress in b with the
counter d in action is less simple than it would be if d and
e were collinear; but the member b will have its true maxi-
mum when the member d does not act, the train approaching
from the left.

GRAPHIC STATICS.

1 60. Influence Lines. — From the foregoing discussion it is
seen that the influence lines for the main truss members may be
drawn by the general method given in Art. 147. In one case,
however, there is a peculiarity which is worthy of special notice.

Considering the lower chord member C1D1 (Fig. 49, PL VIII),.
the point Fl falls between Al and Clt making nl negative. The

ordinate F2f2 is made equal to — A positive ; A2f2 and f2B2 are

drawn, intersecting verticals through C2 and D2 in CB and DB,
respectively ; the influence line is A2C^DSB2. This peculiar
form does not occur in any common form of truss without
subordinate bracing.

Figure 50 shows the influence line for the upper chord mem-
ber^-; the subordinate braces being so placed that the diago-
nals are in tension. Here n2 is negative, F1 falling between
D^ and Bv

For the subordinate members the influence lines may be
drawn without difficulty from the principles of Art. 158.

161. Application. — On PL VII is shown a truss of 300 ft
span, divided into ten panels of 30 ft. each. The lengths of the
main verticals are 30 ft., 45 ft., and 60 ft. The subordinate
diagonals are so placed as to sustain tension. Counterbraces
are shown in the central panel only ; a full computation of
stresses due to live and dead loads will determine whether other
counters are needed. The moving load is the same as shown
on PL III.

At (A), PL VII, is shown the position of the truss causing
greatest stress in the upper chord member C^F^, and at (B) the
position causing greatest stress in the web member C^D^. The
letters A, B, C, D, F have the same meanings as on Plates V
and VI, and the construction is self-explanatory. If the lower
part of the web member C^K, in Fig. (B\ were under consid-
eration, the point Fl would be as shown in the figure, but ClDl
would be the double panel C^K (the same as if the subordinate

UNIFORMLY DISTRIBUTED MOVING LOAD. 145

members were omitted). These statements are in accordance
with the foregoing discussion.

In each of the cases shown on PI. VII it will be seen that the
lines A1 B* and CD* intersect at a point F1 lying in the vertical
through F^

The stresses corresponding to the positions thus determined
may be computed by the method of moments, as on PL VI.
The construction is not shown ; but satisfactory results may be
secured by careful work and by the use of suitable linear and
force scales.

The construction for determining the position for greatest
stress in a subordinate member is not shown ; it is identical with
the construction shown on PL IV for determining the position
for greatest bending moment at a given section of a beam.

§ 5. Uniformly Distributed Moving Load.

162. Application of General Method. — The case in which the
moving load is assumed to have a uniform horizontal distribu-
tion involves no principles additional to those already explained.
The foregoing discussion applies to any load distribution what-
ever. It is, however, of interest to notice the special form
assumed by the general principles when the load is uniformly
distributed.

The load polygon becomes a straight line whose slope depends
upon the intensity of the loading (the load per unit length).
The funicular polygon becomes a continuous curve, being in
fact a parabola with axis vertical, the position of the vertex
depending upon the position of the pole of the force diagram.
For the purpose of drawing the funicular polygon the load may
be replaced by a series of concentrated loads ; the sides of the
resulting polygon will be tangent to the desired curve, and the
polygon will coincide more and more nearly with the desired
parabola, the smaller and nearer together the concentrated loads
are taken.

146 GRAPHIC STATICS.

It is obvious that for maximum chord stresses the only condi-
tion is that the moving load shall cover the whole span.

For maximum stress in any web member, the head of the
load series must be at some point in the corresponding panel ;
the exact position may be determined by applying the general
method. The position of the head of the load series is shown
at a glance by the influence diagram. Thus, from diagram (£*),
Fig. 48 (PI. VIII), it is obvious that the greatest tension in the
member C^D^ will occur when the load covers E±Bl completely,
E^A^ being free from load.

163. Assumption of Full Panel Loads. — The determination
of the maximum web stresses is somewhat simplified by an
approximate assumption as to the way in which the uniformly
distributed load comes upon the truss. Thus, in Fig. 53, let it
be required to determine the greatest tension in the member
BB1 '. The moving load must be brought on from the right until
its head is at some point between A and B. In this position,
the loads actually supported at the various joints are as follows:

A' B' c'

X A B C" C Z D E F Y

Fig. 53

At C a full panel load (half the load between B and C, and half
that between C and D} ; at each of the points D, E, and F also
a full panel load ; at B less than a full panel load (half the load
on BC, and a portion of that on AB). Let it be assumed that,
when the tension in BB] is greatest, all joints from B to F inclu-
sive sustain full panel loads, while all joints to the left of B
(only one in the case shown) are free from loads.

If a similar assumption is made in case of every chord mem-
ber, the resulting stresses will not differ greatly from those
obtained by rigorously adhering to the assumption of a uniformly
distributed load, the error being on the side of safety. It is to

UNIFORMLY DISTRIBUTED MOVING LOAD. 147

be remarked, also, that the case of a strictly uniform dis-
tribution is not realized in practice, so that the results of
the above assumption are probably as reliable as would be
obtained by following out strictly the assumption of uniform
distribution.

The method of determining maximum stresses, on the above
assumption as to the loading, will be illustrated by reference to
the truss shown in Fig. 54, PL VIII. In this truss the panels
are of equal length ; but the method applies also to the case of
unequal panels, the only difference being that the " full panel
loads " for different joints are not equal unless the panels are
equal.

The principle of counterbracing will be employed here, the
diagonals being constructed to sustain tension only.

164. Dead Load Stresses. — The dead load stresses may
be determined by means of a stress diagram, as in the roof-
truss problems already treated. Two points should be ob-
served in drawing this diagram, (i) If dead loads are taken
to act at upper as well as at lower joints, the force poly-
gon for the loads and reactions must show these forces in
the same order as that in which their points of application
occur in the perimeter of the truss. (2) The diagonal mem-
bers assumed to act are taken as all sloping in the same
direction.

The reason for the first point is the same as already explained
in Art. 90; viz. that unless the forces be taken in the order
mentioned, the stress diagram cannot be the true reciprocal of
the truss diagram and certain lines will have to be duplicated.
The reason for assuming the diagonals as all sloping in the
same way is the same as in the case of the roof truss with
counterbracing (Arts, in and 112).

The dead-load stress diagram is not shown, since its con-
struction involves no principle not already fully explained and
illustrated.

I48 GRAPHIC STATICS.

165. Stresses Due to Moving Load. — (a) Chord members. —
By applying the method of sections it is easily seen that a load
at any point produces tension in every lower chord member and
compression in every upper chord member. It follows that the
greatest stresses in the chord members will occur when the truss
is fully loaded. A convenient method of determining the live
load stresses is, therefore, to draw a stress diagram, as in case of
fixed loads. This diagram is not shown.

(b) Web members. — In case of a web member, loads in dif-
ferent positions tend to cause opposite kinds of stress. Thus,
considering the member f'n' (Fig. 54, PL VIII), a tension is
caused in it by a load at either of the points ab, be, cd, de, or ef,
while a compression is caused by a load at fg, gh, or hi. Simi-
larly, loads at ab, be, cd, de, and ef all tend to throw compression
on the vertical member f'm' , while loads on fg, gh, hi have the
opposite tendency.

Therefore, to produce the greatest tension upon f'n1 (and
compression on/W), the live load must act only at ef and all
joints to the right; while to cause the greatest compression on
f'n' (and tension upon f'm') the live load must act only at fg,
gh, and hi. A similar statement will hold regarding any other
web member. Since counterbraces are to be used in all panels
in which diagonal members would otherwise be thrown into
compression, we shall need only to consider the greatest tension
in each diagonal and the greatest compression in each vertical.
We shall first outline the method to be employed, and then
explain the construction.

To determine the greatest tension in a diagonal member, as
g1m' : Assume the live load to come upon the bridge from the
right until there are full loads at the joints ab, be, cd, de, ef, and
fg. Take a section cutting g'm' and the two chord members
gg' and m'm, and consider the forces acting upon the portion of
the truss to the left of the section. These forces are four in
number : the reaction at the support and the forces acting in the
three members cut. Hence we first determine the reaction,

UNIFORMLY DISTRIBUTED MOVING LOAD. 149

and then determine the three other forces for equilibrium by
the method of Art. 42.

The construction is shown in Fig. 54 (PI. VIII). ABCDEFGHI
is the force polygon for the eight live loads that may come
upon the truss. Choosing a pole O, the funicular polygon
for the eight loads is next drawn. Now, turning the attention
to the member g'm', the loads gh and ///are assumed not to act.
The reactions at the supports for this case of loading are found
in the usual way. Prolong oa and og to intersect the lines
of action of the two reactions, and join the two points thus
determined. This gives the closing line of the funicular poly-
gon (or oni). The ray OM is now drawn parallel to the string
om, and the two reactions are GM and MA.

Now take a section through mm', m'g1, and g'g, and apply
the construction of Art. 42 to the determination of the forces
acting in the three members cut. The resultant of GM and
the force in gg1 must act through the point X (the intersection
of gg1 produced and ij\ The resultant of the forces in ;;/;;/'
and m'g1 must act through their intersection Y. Hence these
two resultants (being in equilibrium with each other) must both
act in the line XY. From G draw a line parallel to gg1 , and
from J/a line parallel to XY', mark their point of intersection
G1 . Then MG' is the resultant of the forces in mm' and m'g1 .
From M draw a line parallel to mm', and from G1 a line parallel
to the member g'm', and mark their point of intersection M' .
Then M'G' represents the force in the member m'g1. This is
also the value of the greatest stress in m'g' .

To determine the stress in the vertical member I'g1, the
same loading must be assumed, and a similar construction is
employed. Take a section cutting II', I'g', and^^f, and deter-
mine forces acting in these three lines which shall be in equi-
librium with the reaction GM. This reaction is in equilibrium
with MG' and G' G, the former having the line of action XY.
Then MG' is resolved into two forces having the directions of
the members g' I' and I' I. The stress in g'l' is found to be

150 GRAPHIC STATICS.

a compression, represented in the stress diagram by the line
G'Lf.

The maximum live load stress in every web member may be
found in the same way. If the above reasoning is understood,
there will be no difficulty in applying the same method to the
remaining members.

1 66. Maximum Stresses. — By combining the stresses due to
live and dead loads, the maximum stresses are easily deter-
mined.

Web members. — When the web members are considered, the
effect of counterbracing needs careful attention.

The construction above explained gives the greatest live load
tension in each diagonal. It may be that for certain members,
this tension is wholly counterbalanced by the dead loads. In
any panel in which this is the case, the member shown will
never act and may be omitted. The counterbrace must then
be considered.

Evidently, the algebraic sum of the stresses due to live and
dead loads will be the true maximum tension in each of the
diagonals shown. For the greatest tension in the other system
of diagonals, the load must be brought on the bridge from the
left; or, what amounts to the same thing, the tension already
found for any one of the diagonals shown is also the greatest
tension in the diagonal sloping the opposite way in the panel
equally distant from the middle of the truss. (In fact, the
stresses in all members due to a movement of the loads from
left to right may be obtained from the results already reached,
by consideration of symmetry.)

As to the vertical members, two values of the stress must be
compared in every case — namely, the greatest compressions
corresponding to the two directions of the moving load. But
both can be obtained from the above results by considering the
symmetry of the truss. For example, the stress found for I'g3
is a possible value for the stress in r'b' , and must be compared

UNIFORMLY DISTRIBUTED MOVING LOAD. 151

with the value obtained for the latter member when the load
moves from right to left. It is possible, also, that certain of
the verticals may be in tension when the dead loads act
alone.

Maximum chord stresses. — These are found by combining
the stresses due to fixed and moving loads, determined as.
already described.

PART III.

CENTROIDS AND MOMENTS OF INERTIA.

CHAPTER VIII. CENTROIDS.

§ i. Centroid of Parallel Forces.

167. Composition of Parallel Forces. — The composition of
complanar parallel forces can always be effected by means
of the funicular polygon, by the method of Art. 27. It is now
necessary to consider parallel systems more at length, as a pre-
liminary to the discussion of graphic methods for determining
centers of gravity and moments of inertia.

168. Resultant of Two Parallel Forces. — Let ab and be (Fig.
55) be the lines of action of two parallel forces, their magni-

tudes being AB and BC (not shown). Let
ac be the line of action of their resultant,
and AC (not shown) its magnitude. By
the principle of moments (Art. 50) the sum
of the moments of AB and BC about any
point in their plane is equal to the moment
of A C about the same point. If the origin of moments is on
.ac, the moment of AC is zero; and therefore the moments of
AB and BC are numerically equal (but of opposite signs).

Let any line be drawn perpendicular to the given forces,
intersecting their lines of action in Pf, Q!, and R' respectively.

g. 55

'52

CENTROID OF PARALLEL FORCES.

153

Let any other line be drawn intersecting the three lines of
action in P, Q, and R respectively. Then

PR QR
P'R' Q'R1'

and therefore ABxPR=BCx QR.

That is, PQ is divided by the line ac into segments inversely

proportional to AB and BC.

If AB and BC act in opposite directions, the line ac will be
outside the space included between ab and be ; but the above
result is true for either case.

169. Centroid of Two Parallel Forces. — If the lines of action
ab and be (Fig. 55) be turned through any angle about the points
P and Q respectively, the forces remaining parallel and of
unchanged magnitudes, the line of action of their resultant will
always pass through the point R. For, by the preceding article,
the line of action of the resultant will always intersect PQ in a
point which divides PQ into segments inversely proportional to
AB and BC. Hence, if AB and BC remain unchanged, and
also the points P and Q, the point R must also remain fixed,

If P and Q are taken as the points of application of AB and
BC, R may be taken as the point of application of AC, in
whatever direction the parallel forces are supposed to act. The
point R is called the centroid * of the parallel forces AB and
BC for the fixed points of application P and Q.

170. Centroid of Any Number of Parallel Forces. — Let AB,
BC, and CD be three parallel forces, and let P, Q, and S
(Fig. 56) be their fixed points of

application. Let R be the cen-
troid of AB and BC, and A C their
resultant. Take R as the fixed
point of application of AC, and
determine T, the centroid of AC FiS. so

* The name center of parallel forces has been quite commonly used instead of centroid
as above denned. The latter term has, however, been adopted by some of the later writers,
and seems on the whole a better designation.

!54 GRAPHIC STATICS.

and CD. Let AD be the resultant of AC and CD ; then AD is
also the resultant of AB, BC, and CD.

Now if ABt BC, and CD have their direction changed, but
still remain parallel and unchanged in magnitude, it is evident
that the point T, determined as above, will remain fixed and will
always be on ad, the line of action of AD. The point T is called
the centroid of the three forces AB, BC, and CD.

By an extension of the same method, a centroid may be
determined for any system of parallel forces having fixed points
of application. Hence the following definition may be stated :

The centroid of a system of parallel forces having fixed points
of application is a point through which the line of action of
their resultant passes, in whatever direction the forces be taken
to act.

In determining the centroid by the method just described,
the forces may be taken in any order without changing the
result. For the centroid must lie on the line of action of the
resultant ; and since this is a determinate line for each direction
in which the forces may be taken to act, there can be but one
centroid.

171. Non-complanar Parallel Forces. — The reasoning of the
preceding articles is equally true, whether the forces are corn-
planar or not. In what follows we shall deal either with
complanar forces, or with forces whose points of application are
complanar. No more general case will be discussed.

172. Graphic Determination of Centroid of Parallel Forces. —

If the line of action of the resultant of any system of parallel
forces be found for each of two assumed directions of the forces,
the point of intersection of these two lines is the centroid of
the system. Moreover, if the points of application are com-
planar, the two assumed directions may both be such that the
forces will be complanar.

Thus, let the plane of the paper be the plane containing the
given points of application, and let ab, be, cd, de, ef (Fig. 57) be

CENTROID OF PARALLEL FORCES.

155

the points of application of five parallel forces, AB, BC, CD,
DE, EF. Draw through these points parallel lines in some
chosen direction, and taking them as the lines of action of the
given forces, construct the force and funicular polygons corre-

sponding. The line of action of the resultant is drawn through
the intersection of the strings oa and of, and this line contains
the centroid of the given forces. Next draw through the given
points of application another set of parallel lines, preferably per-
pendicular to the set first drawn, and draw a funicular polygon
for the given forces with these lines of action. (It is unneces-
sary to draw a new force diagram, since the strings in the
second funicular polygon may be drawn respectively perpendic-
ular to those of the first.) This construction determines a
second line as the line of action of the resultant corresponding
to the second direction of the forces. The required centroid of
the system is the point of intersection of the two lines of
action of the resultant thus determined, and is the point of in
the figure.

Example. — Find graphically the centroid of the following
system of parallel forces, and test the result by algebraic com-
putation : A force of 20 Ibs. applied at a point whose rectangu-
lar coordinates are (4, 6) ; 12 Ibs. at the point (12, 3) ; 20 Ibs. at

156 GRAPHIC STATICS.

the point (10, 10) ; — 10 Ibs. at the point (7, —9); —8 Ibs. at
the point (—5, — 10).

173. Centroid of a Couple. — If AB and BC are the magni-
tudes of any two parallel forces applied at points P and Q,
then by Art. 169 their centroid R is on the line PQ and is
determined by the equation

PR = BC
QR AB'

If AB and BC are equal and opposite forces, PR and QR must
be numerically equal, and R must be outside the space between
the lines of action of AB and BC. These conditions can be
satisfied only by making PR and QR infinite. Hence we may
say that the centroid of two equal and opposite forces lies on
the line joining their points of application and is infinitely dis-
tant from these points.

174. Centroid of a System whose Resultant is a Couple. —

If a system with fixed points of application is equivalent to a
couple, its centroid will be infinitely distant from the given
points of application. A line containing this centroid can be
determined as follows :

Take the given forces in two groups ; the resultants of the
two groups will be equal and opposite. Find the centroid of
each group, and suppose each partial resultant applied at the
corresponding centroid. Then the centroid of the whole sys-
tem is the same as that of the couple thus formed, and will lie
on the line joining the two partial centroids.

If the separation into groups be made in different ways, dif-
ferent couples and different partial centroids will be found.
The different couples are, of course, equivalent ; and it may be
proved that the lines joining the different pairs of partial cen-
troids are all parallel, and intersect in the (infinitely distant)
centroid of the whole system.

For, suppose the given system to be equivalent to a couple

CENTER OF GRAVITY. 157

Q with points of application A and B, and also to a couple Qf
with points of application A1 and B1. These two couples must
be equivalent to each other, whatever be the direction of the
forces. Let AB be taken as this direction. The two equal
and opposite forces of the couple Q neutralize each other, since
their lines of action are coincident ; hence the two forces of
the couple Q must also neutralize each other. Therefore A'B'
must be parallel to AB.

175. Moment of a Force about an Axis. — The moment of a
force with respect to a given axis, as defined in Art. 47,
depends not only upon the point of application of the force,
but upon its direction. In dealing with systems of forces
whose direction may change but whose points of application
are complanar, we shall need to compute moments only for
axes lying in the plane of the points of application ; and the
forces may usually be regarded as acting in lines perpendicular
to this plane. Hence we shall compute moments by the
following rule :

The moment of a force with reference to any axis is the
product of the magnitude of the force into the distance of its
point of application from the* axis.

§ 2. Center of Gravity — Definitions and General Principles.

176. Center of Gravity of Any Body. — Every particle of
a terrestrial body is attracted by the earth with a force propor-
tional directly to the mass of the particle. The resultant of
such forces upon all the particles of a body is its weight ; and
the point of application of this resultant is called the center of
gravity of the body. The lines of action of these forces may
be assumed parallel without appreciable error. We may there-
fore define the center of gravity of a body as follows :

If forces be supposed to act in the same direction upon all
particles of a body, each force being proportional to the mass

158 GRAPHIC STATICS.

of the particle upon which it acts, the centroid of this system
of parallel forces is the center of gravity of the body.

This point is also called center of mass, and center of inertia,
either of which is a better designation than center of gravity.
The latter term is, however, in more general use.

177. Centers of Gravity of Areas and Lines. — The term
center of gravity, as above denned, has no meaning when
applied to lines and areas, since these magnitudes have no
mass, and hence are not acted upon by the force of gravity.
It is, however, common to use the name center of gravity in
the case of lines and areas, with meanings which may be stated
as follows :

The center of gravity of an area is the center of gravity of
its mass, on the supposition that each superficial element has a
mass proportional to its area. This point would be better
described as the center of area.

The center of gravity of a line is the center of gravity of its
mass, on the assumption that each linear element has a mass
proportional to its length. The term center of length is pref-
erable, and will often be used in what follows.

Similar statements might be made regarding geometrical
solids, but we shall have to deal chiefly with lines and areas.

178. Moments of Areas and Lines. — Definition. — The mo-
ment of a plane area with reference to an axis lying in its plane
is the product of the area by the distance of its center of
gravity from the axis.

Proposition. — The moment of any area about a given axis is
equal to the sum of the moments of any set of partial areas
into which it may be divided. For, by the definition of center
of gravity, a force numerically equal to the total area and
applied at its center of gravity is the resultant of a system of
forces numerically equal to the partial areas and applied at
their respective centers of gravity ; and the moment of any

CENTER OF GRAVITY. 159

force is equal to the sum of the moments of its components
(Art 50).

A similar definition and proposition may be stated regarding
lines.

The moment of an area or line is zero for any axis containing
its center of gravity.

179. Symmetry. — Two points are symmetrically situated
with respect to a third point if the line joining them is bisected
by that point.

Two points are symmetrically situated with respect to a line
or plane when the line joining them is perpendicular to the
given line or plane and bisected by it.

A body is symmetrical with respect to a point, a line, or a
plane, if for every point in the body there is another such that
the two are symmetrically situated with respect to the given
point, line, or plane. The point, line, or plane is in this case
called a point of symmetry, an axis of symmetry, or a plane of
symmetry of the body.

180. General Principles. — (i) The center of gravity of two
masses taken together is on the line joining the centers of
gravity of the separate masses. For it is the point of applica-
tion of the resultant of two parallel forces applied at those
points.

(2) If a body of uniform density has a plane of symmetry,
the center of gravity lies in this plane. If there is an axis of
symmetry, the center of gravity lies in this axis. If the body
is symmetrical with respect to a point, that point is the center
of gravity. For the elementary portions of the body may be
taken in pairs such that for each pair the center of gravity is
in the plane, axis, or point of symmetry.

1 8 1. Centroid. — The center of gravity of any body or geo-
metrical magnitude is by definition the same as the centroid of
a certain system of parallel forces. It will be convenient,

160 GRAPHIC STATICS.

therefore, to use the word centroid in most cases instead of
center of gravity.

§ 3. Centroids of Lines and of Areas.

182. General Method of Finding Centroid. — The centroid
of any area may be found by the following method : Divide the
given area into parts such that the area and centroid of each
part are known. Take the centroids of the partial areas us the
points of application of forces proportional respectively to those
areas. The centroid of this system of forces is the centroid of
the total area, and may be found by the method of Art. 172.

The centroid of a line may be found by a similar method.

The method just described is exact if the magnitudes of the
partial areas and their centroids are accurately known. If the
given area is such that it cannot be divided into known parts, it
will still be possible to get an approximate result by this
method.

In applying this general method, it is frequently necessary to
know the centroids of certain geometrical lines and figures, and
also the relative magnitudes of the areas of such figures.
Methods of determining these will be given in the following
articles.

183. Centroids of Lines. — The centroid of a straight line is
at its middle point.

Broken line. — The centroid of a broken line is the center of
a system of parallel forces, of magnitudes proportional to the
lengths of the straight portions of the line, and applied respec-
tively at their middle points. It may be found graphically by
the method of Art. 172, or by any other method applicable to
parallel forces.

Part of regular polygon. — For the centroid of a part of a
regular polygon, a special construction is found useful.

CENTROIDS OF LINES AND OF AREAS. 161

Let ABCDE (Fig. 58) be part of such a polygon, and O the
center of the inscribed circle. Let
r = radius of inscribed circle ; / =
length of a side of the polygon ; s
= total length of the broken line
AE. Through O draw OC, the axis
of symmetry of AE; and MN, per-
pendicular to OC.
First, the centroid must lie on OC.

Second, to find its distance from O, assume a system of equal
and parallel forces applied at the middle points of the sides AB,
BC, etc. The required centroid is the point of application of
the resultant of these forces. Taking MN as the axis of
moments, and letting x = required distance of centroid from
MN, and x^ x^, xz, x± the distances of the middle points of AB,
BC, etc., from MN, we have from the principle of moments,

IX-L 4- lxz 4- &s 4- /-** = sx.

But if Ab and Bb be drawn perpendicular respectively to PQ
and MNwe have from the similar triangles ABb, POQf

= _ _=^
Ab PQ°T Ab xi

In the same way,

lxz = r* be,

Hence, s-x=r (Ab + bc+cd+dE) = r> AE,

where AE is equal to the projection of the broken line ABCDE
onMN.

The centroid G may now be found graphically as follows :
Make Oc' = r\ ON=%s\ OE' = % AE', draw Nc'. Then G is
determined by drawing E'G parallel to Nc'

162 GRAPHIC STATICS.

Circular arc. — The above construction holds, whatever the
length of the side /. If this length be decreased indefinitely,
while the number of sides is increased indefinitely, so that the
length s remains finite, we reach as the limiting case a circular
arc. The same construction therefore applies to the determina-
tion of the centroid of such an arc, r denoting the radius of the
circle and s the length of the arc.

184. Centroids of Geometrical Areas. — Parallelogram. — The
centroid of a parallelogram is on a line bisecting two opposite
sides.

Let ABCD (Fig. 59) be a parallelogram, and EF a line
bisecting AD and BC. Divide AB into any even number of

equal parts, and through the points
of division draw lines parallel to BC.
Also divide fiCinto any even number
of equal parts and draw through the
points of division lines parallel to
AB. The given parallelogram is thus
divided into equal elements. Now
consider a pair of these elements, such as those marked P and
Q in the figure, equally distant from AD, and also equally
distant from EF, but on opposite sides of it. The centroid of
the two elements taken together is at the middle point of the
line joining their separate centroids. If the number of divisions
of AB and of BC be increased without limit, the elements
approach zero in area, and the centroids of P and Q evidently
approach two points which are equally distant from EF. Hence
in the limit, the centroid of such a pair of elements lies on the
line EF. But the whole area ABCD is made up of such pairs ;
hence the centroid of the whole area is on the line EF. For
like reasons it is also on the line bisecting AB and DC] hence
it is at the intersection of the two bisectors.

The point thus determined evidently coincides with the point
of intersection of the diagonals A C and BD.

CENTROIDS OF LINES AND OF AREAS.

163

Triangle. — The centroid of a triangle lies on a line drawn
from any vertex to the middle of the opposite side ; and is,
therefore, the point of intersection of the three such lines.

Let ABC (Fig. 60) be any triangle, and D the middle point
of BC. Then the centroid of ABC must lie on AD. For AD
bisects all lines, such as be, parallel to BC. Now inscribe in
the triangle any number of parallelograms such as bcc'V, with
sides parallel respectively to BC and
AD. The centroid of each parallelo-
gram lies on AD, and, therefore, so
also does the centroid of the whole
area composed of such parallelograms.
If the number of such parallelograms
be increased without limit, the alti-
tude of each being diminished without limit, their combined
area will approach that of the triangle, and the centroid of this
area will approach in position that of the triangle. But since
the former point is always on the line AD, its limiting position
must be on that line. Therefore the line AD contains the cen-
troid of the triangle.

By the same reasoning, it follows that the centroid of ABC
must lie on BE, drawn from B to the middle point of AC.
Hence it must be the point of intersection of AD and BE,
which point must also lie on the line CF drawn from C to the
middle point of AB.

The point G divides each bisector into segments which are
to each other as I to 2. For, from the similar triangles ABC,
EDC, since EC is half of AC, it follows that DE is equal to
half of BA. And from the similar triangles AGB, DGE, since
DE is half of AB, it follows that GE is half of GB, and GD
half of GA.

Quadrilateral — Let ABCD (Fig. 61) be a quadrilateral of
which it is required to find the center of gravity. Draw BD,
and let E be its middle point. Make EG\=-\ EA, and EG^ = \
EC. Then the centroids of the triangles ABD and BCD are

1 64

GRAPHIC STATICS.

Fig. 61

Gl and G2 respectively. Hence the centroid of ABCD is on
the line GiGa at a point dividing it into segments inversely

proportional to the areas of ABD
and BCD. Since these two tri-
angles have a common base, their
areas are proportional to their alti-
tudes measured from this base.
But these altitudes are propor-
tional to AF and FC, or to G^H and GJJ ; hence, if G is the
required centroid,

G\G '. G%G '. '. GiH '. G\H .

Therefore G is found by making G^G=G^H.

Circular sector. — To find the centroid of a circular sector
OAB (Fig. 62) we may reason as follows : Draw two radii OM>
ON, very near together. Then OMN differs little from a tri-
angle, and its centroid will fall very near the arc AB\ drawn
with radius equal to two-thirds of OA. If the whole sector be
subdivided into elements such as OMN, their centers of gravity
will all fall very near to the arc A'£'. If the
number of such elements is indefinitely in-
creased, the line joining their centroids
approaches as a limit the arc A'£'. And
since the areas of the elements are propor-
tional to the lengths of the corresponding
portions of AB\ the centroid of the total
area is the same as that of the arc A'B'.
This point may be found by the method described in Art. 183.

185. Graphic Determination of Areas. — Let there be given
any number of geometrical figures, and let it be required to
determine the relative magnitudes of their areas.

If a number of rectangles can be found, of areas equal
respectively to the areas of the given figures and having one
common side, then the remaining sides will be proportional to
the areas of the given figures.

M N

CENTROIDS OF LINES AND OF AREAS.

I6S

An important case is that in which the given figures are
such that the area of each is equal to the product of two known
lines. In this case a series of equivalent rectangles having one
common side can be found by the following construction.

Let ABC (Fig. 63) be a triangle, and let it be required to
determine an equivalent
rectangle having a side of
given length as LM. Let
b and h be the base and
altitude of the triangle.
Make LN= |- b and LP = //,
and draw the semicircumference PRN,
perpendicular to PN\ then

LI? = PLx LN= J bh = area ABC.
Draw MR, and perpendicular to it draw RQ. Then

ABC.

N M

Fig. 63

Draw the ordinate LR

Hence LQ is the required length.

If the given figure is a parallelogram, LN and LP may be
its base and altitude. If it be a circular sector, LN and LP
may be the length of the arc and half the radius.

1 86. Centroids of Partial Areas. — It may be required to
find the centroid of the part remaining after deducting known
parts from a given area. For this case the construction of
Art. 182 needs modification. In the case there considered we
regarded the partial areas and the total area as proportional to
parallel forces acting at their respective centers of gravity. In
this case we may also represent the total area, the portions
deducted from it, and the remaining portion as forces acting at
the respective centers of gravity ; but the forces corresponding
to the areas deducted must be taken as acting in the opposite
direction to that assumed for the forces representing the total
area and the area remaining.

Thus, to find the centroid of the area remaining after deduct-

1 66

GRAPHIC STATICS.

ing from the circle ABD (Fig. 64) a smaller circle EFH and a
sector OAB, we may proceed as follows : Find the centroid of
three parallel forces proportional to the
areas ABD, EFH, and OAB, and applied
at their respective centroids O, C, and G ;
but the last two must be taken as acting
in the direction opposite to that of the first.
With this understanding, the force and
funicular polygons may be employed as in
Art. 172.

187. Moments of Areas. — The moment of an area about any
line in its plane may be determined from the funicular polygon
employed in finding its center of gravity. Let the parallel
forces applied at the centroids of the partial areas be assumed
to act parallel to the axis of moments. Then the distance inter-
cepted on the axis by the extreme lines of the funicular poly-
gon, multiplied by the pole distance, is equal to the moment of
the total area about the axis. For, by Art. 56, this construc-
tion gives the moment of the resultant force about any point
in the given axis ; and this is equal to the moment of the result-
ant area about the axis by definition.

A similar rule gives the moments of the partial areas.

CHAPTER IX. MOMENTS OF INERTIA.
§ I. Moments of Inertia of Forces.

1 88. Definitions. — The moment of inertia of a body with
respect to an axis is the sum of the products obtained by mul-
tiplying the mass of every elementary portion of the body by
the square of its distance from the given axis.

The moment of inertia of an area with respect to an axis is
the sum of the products obtained by multiplying each element-
ary area by the square of its distance from the axis.

The moment of inertia of a line may be similarly defined,
using elements of length instead of elements of mass or area.

The moment of inertia of a force with respect to any axis is
the product of the magnitude of the force into the square of the
distance of its point of application from the axis. The sum of
such products for any system of parallel forces is the moment
of inertia of the system with respect to the given axis.

[NOTE. — The term moment of inertia had reference originally to material bodies,
the quantity thus designated having especial significance in dynamical problems relat-
ing to the rotation of rigid bodies. The quantity above defined as the moment of
inertia of an area is of frequent occurrence in the discussion of beams, columns, and
shafts in the mechanics of materials. In the graphic discussion of moments of iner-
tia of areas, it is convenient to treat areas as forces, just as in the determination of
centers of gravity; it is therefore convenient to use the term moment of inertia of a
force in the sense above defined. It is only in the case of masses that the term
moment of inertia is really appropriate, but it is by analogy convenient to apply it
to the other cases.]

The product of inertia of a mass with respect to two planes is
the sum of the products obtained by multiplying each element-
ary mass by the product of its distances from the two planes.

167

168 GRAPHIC STATICS.

The product of inertia of an area, a line, or a force may be
•defined in a similar manner.

For a plane area, the product of inertia with respect to
two lines in its plane may be defined as the sum of the
products obtained by multiplying each element of area by the
product of its distances from the given lines. This is equiv-
alent to the product of inertia of the area with respect to two
planes perpendicular to the area and containing the two given
lines.

For a system of forces with points of application in the same
plane, the product of inertia with reference to two axes in that
plane may be defined as the sum of the products obtained by
multiplying the magnitude of each force by the product of the
distances of its point of application from the given axes. It is
with such systems and with plane areas that the following pages
chiefly deal.

The radius of gyration of a body with respect to an axis is the
distance from the axis of a point at which, if the whole mass of
the body were concentrated, its moment of inertia would be
unchanged. The square of the radius of gyration is equal to
the quotient obtained by dividing the moment of inertia of the
body by its mass.

The radius of gyration of an area may be defined in a similar
manner.

The radius of gyration of a system of parallel forces is
the distance from the axis of the point at which a force
equal in magnitude to their resultant must act in order that
its moment of inertia may be the same as that of the system.
The square of the radius of gyration may be found by dividing
the moment of inertia of the system by the resultant of the
forces.

189. Algebraic Expressions for Moment and Product of In-
ertia.— Moment of inertia. — Let m^ mz, etc., represent ele-
mentary masses of a body, and r^ r2, etc., their respective

MOMENTS OF INERTIA OF FORCES. 169

distances from an axis ; then the moment of inertia of the body
with respect to that axis is

m\r? + m^r} -\ ---- = 2 mr2,

the symbol 2 being a sign of summation, and the second mem-
ber of the equation being merely an abbreviated expression for
the first.

Product of inertia. — Let/i,/2, etc., denote the perpendicular
distances of elements m\t mz, etc., from one plane, and q^ qz, etc.,
their distances from another ; then the product of inertia of the
body with respect to the two planes is

m\p\q\ + mzpzqz H — = ^mfq.

Radius of gyration. — With the same notation, if k denotes
the radius of gyration of the body, we have

(m: + m2-\ — ) & = nur? -f mzrz -\ — .
Or,

72 _ \ ,z -\

ml + m2-\ ---- *#

Here 2m is equal to the whole mass of the body.

Product-radius. — Let c represent a quantity defined by the
equation.

(mi + mz + • • •) c* — Wip-fli + m^p2qz -\ ---- .

Then if the whole mass were concentrated at the same distance
c from two axes, its product of inertia with respect to those
axes would be unchanged. This quantity c is thus seen to be
analogous to the radius of gyration. It may be called the
product-radius of the body with respect to the two axes. The
value of c is always given by the equation

Expressions similar to those just given apply also to plane
areas and to systems of parallel forces. In case of an area,
*#i, mz, etc., denote elements of area; r^ rZy etc., their distances
from the axis of inertia ; and (pi, q^, (/2, ^2), etc., their dis-

UNIVERSI

GRAPHIC STATICS.

tances from the two planes. In case of a system of parallel
forces with complanar points of application, nil, mz, etc., must
be replaced by the magnitudes of the forces.

190. Determination of Moment of Inertia of a System of
Parallel Forces. — Let the points of application of the forces
be in the same plane, which also contains the assumed axis.
We shall have to deal only with systems satisfying these condi-
tions. By the definition, the moment of inertia will be the
same, whatever the direction of the forces. If we take the
moment of any force (as defined in Art. 175) about the given
axis, and suppose a force equal in magnitude to this moment
to act at the point of application of the original force, and in a
direction corresponding to the sign of the moment, then the
moment of this new force about the given axis is equal to
the moment of inertia of the original force. If this be done for
all the forces, the algebraic sum of the results will be the
required moment of inertia of the system.

This process can be carried out graphically by methods
already described.

Let ab, be, cd, de (Fig. 65) be the points of application of
four parallel forces, and let the axis of inertia be QR. Suppose
all the forces to act in lines parallel to QR, passing through the
given points of application. Their respective moments with
reference to QR may now be found by the method of Art. 55.

Draw the force polygon ABCDE and choose a pole O, taking
the pole distance H preferably equal to AE or some simple
multiple of AE. (In Fig. 65 H is taken equal to AE.) Draw
a funicular polygon and prolong each string to intersect QR.
Then the moment of any force with respect to QR is the
product of H by the distance intercepted on QR by the two
strings corresponding to the force in question. Thus the
moment of AB is given by the distance A' B' (Fig. 65) multi-
plied by H. Also the successive moments of BC, CD, DE are
represented by £'Cr, C'D', £>'£', each multiplied by H. It is

MOMENTS OF INERTIA OF FORCES.

I7T

seen also that the intercepts, if read in the above order, give a
distinction between positive and negative moments ; upward
distances on QR denoting in this case positive moments, and
downward distances negative moments.

We have now to find the sum of the moments of a second
system of forces acting in the original lines, but represented in

a l> be

Fig.

magnitude and direction by the intercepts just found. We
may take as the force polygon for the second construction the
line A'B'C'D'E' (Fig. 65), and choosing any pole distance //',
draw a second funicular polygon and find the distances inter-
cepted by the successive strings on the line QR. But in this
case, since only the resultant is desired, we need only find the
intercept An E" between the first and last strings. The product
of this intercept by Hr gives the sum of the moments of A'B',
B'C', C'D', D'E' with respect to QR ; and, if the product 'be
multiplied by H (since A'B', B'C', etc., should each be multi-
plied by H in order to represent the magnitudes of the forces
of the second system), the result will be the required moment of
inertia of the given system of forces AB, BC, CD, DE.

172 GRAPHIC STATICS.

It should be noticed that in Fig. 65 (A) is a force diagram,
(B) a space diagram (Art. n); that is, every line in (A) repre-
sents a force, while every line in (B) represents a distance.
Even A'B', £>'Cf,.etc., though used as forces, are actually
merely distances ; and the moment of any one of them is the
product of a length by a length.

191. Radius of Gyration. — The moment of inertia of the
given system is HxH' xA"E". If H has been taken equal
to AE, the product H' xA"E" must equal the square of the
radius of gyration of the system with respect to QR. The
length of the radius of gyration can be found as follows (Fig.
65) : Draw LN=H' + A"E", and make LM=H'. On ZA^as
a diameter draw a semicircle LPN, and from M draw a line
perpendicular to L N, intersecting the semicircle in P. Then
MP is the length of the required radius of gyration. For by

elementary geometry we have PM — LMxMN.

If His taken equal to nxAE, the moment of inertia is equal
to H' xA"E" xnx AE, and the square of the radius of gyration
is equal to nH'xA"E". Hence in Fig. 65 we should put
either LM=nH't or MN=nxA"£".

192. Central Axis. — If the axis with reference to which the
moment of inertia is found contains the centroid of the given
system of forces, it is called a central axis of the system.

In many cases it is desired to find the moment of inertia with
respect to a central axis whose direction is known while the
position of the centroid is at first unknown. It is to be noticed
that the method shown in Fig. 65 is applicable in this case ;
for the first part of the process is identical with that employed
in finding the centroid of the system. If, in Fig. 65, the
strings oa, oe, of the first funicular polygon be prolonged to
intersect, a line through their point of intersection, parallel to
the direction of the forces, will contain the centroid of the
system. If this line is taken as the inertia-axis, the points A',
B' , C', D', E' are the points in which this axis is intersected by

MOMENTS OF INERTIA OF FORCES. 173

the strings oa, ob, ocy od, oe. No further modification of the
process is necessary.

193. Moment of Inertia Determined from Area of Funicular
Polygon. — In Fig. 65, the moments of the given forces are
represented by the intercepts A'B', B'C', C'D't D'E't each
multiplied by the pole distance H. The moment of inertia of
any force, as AB, is equal to the moment of a force represented
by the corresponding intercept as A'B', supposed to act in the
line ab. Now, by definition (Art. 175) the moment about the
axis QR of a force equal to A'B' acting in the line ab is equal
to double the area of the triangle A' i B' ; hence the moment of
the force HxA'B' is equal to double the area of that triangle
multiplied by//. Similarly, the moment of a force HxB'C',
acting in the line be, is equal to double the area of the triangle
B' 2 C! multiplied by H. Applying the same reasoning to each
force, we see that the sum of the moments of the assumed
forces (HxA'B1, HxB'C', etc.) is equal to 2 //times the sum
of the areas of the triangles A' iB', B' 2C', C' 3 D' , D' ^E'.
In adding these triangles, each must be taken with its proper
sign, corresponding to the sign of the moment represented by
it. Thus, the moments of A'B1, B'C', and D'E1 all have the
same sign, while the moment of C'D' has the opposite sign.
The algebraic sum of the areas is, therefore,

area A1 i .5' + area B' 2 C '-area C' 3 £>' + area D' ^E',

which is equal to the area of the polygon A' i 2 34/^. Hence
this area, multiplied by 2 H, gives the moment of inertia of the
required system of forces.

It may sometimes be convenient to apply this principle in
determining moments of inertia, the area being determined by
use of a planimeter, or by any other convenient method. It
should be noticed that if H is taken equal to the sum of the
given forces (AE), twice the area of the funicular polygon is
equal to the square of the radius of gyration. If H—\ AE, the

174

GRAPHIC STATICS.

square of the radius of gyration is equal to the area of the
polygon.

194. Determination of Product of Inertia of Parallel Forces.

— Assume the points of application of the forces to be in the
plane containing the two axes. If the moment of any force
with respect to one axis be found, and a force equal in magni-
tude to this moment be assumed to act at the point of applica-
tion of the original force, then the moment of this new force
with respect to the second axis is equal to the product of
inertia of the given force for the two axes.

Thus, let abt be, cd, de (Fig. 66) be the points of application
of four parallel forces, and let their product of inertia with
respect to the axes QR, ST be required. Draw ABODE, the

Fig. 60

force polygon for the given forces, assumed to act parallel to
QR. Choose a pole O, the pole distance being preferably
taken equal to AE, or some simple multiple of AE, and draw
the funicular polygon as shown, prolonging the strings to inter-
sect QR in the points A', B', C', D', E'. Now assume a series

MOMENTS OF INERTIA OF FORCES. 175

of forces equal to A'B', B'C', etc., each multiplied by H, to act
at the points ab, be, etc., and determine their moments with
respect to the axis ST. To find these moments, draw lines
through ab, be, etc., parallel to ST, and draw a funicular poly-
gon for the assumed forces taken to act in these lines. The
force polygon is obtained by revolving the line A'fi'C'D'E' until
parallel with 57; and is the line A\ B\ C\ D\ E\ in the figure.
The strings o'a', o'e' of the second funicular polygon intersect
ST in the points A" and £". Hence, calling H' the second
pole distance, the product of inertia of the given system is
equal to Hxff'xA"£".

195. Product-Radius. — If H be taken equal to AE (Fig. 66),
H'xA"E" is equal to the square of the product-radius (Art.
189). Hence the product-radius can be found by a construc-
tion exactly like that employed in finding the radius of gyra-
tion. Thus (Fig. 66) take LM=H' and MN=A"E" \ make
LN the diameter of a semicircle, and draw from J/a line per-
pendicular to LN, intersecting the semicircle in P. Then
MP2=LMx MN = H'xA"E"; hence MP = c, the product-
radius. *

196. Relation between Moments of Inertia for Parallel Axes.

— Proposition. — The moment of inertia of a system of par-
allel forces with reference to any axis is equal to its moment of
inertia with respect to a parallel axis through the centroid of
the system plus the moment of inertia with respect to the given
axis of the resultant applied at the centroid of the system.

Let PI, P2, etc., represent the forces ; x\, x^ etc., the dis-
tances of their points of application from the central axis ; and
a the distance of this central axis from the given axis. Calling
the required moment of inertia A, and the moment of inertia
with respect to the axis through the center of gravity A\ we
have

A=P, (*+*,

GRAPHIC STATICS.

Now P^+P^x2-\ — is the algebraic sum of the moments of
the given forces with respect to the axis through their centroid,,
and is equal to zero ; and P^i + P2x2 2 -\ — =A'. Hence

which proves the proposition.

Radii of gyration. — Let k — radius of gyration of the sys-
tem with respect to the given axis, and k' the radius of gyra-
tion with respect to the central axis, and we have

Hence the equation above deduced may be reduced to the form

197. Products of Inertia with Respect to Parallel Axes. —

Proposition. — The product of inertia of a system of parallel
forces with reference to any two axes is equal to the product
of inertia with reference to a pair of central axes parallel to the
given axes, plus the product of inertia of the resultant (acting
at the centroid) with reference to the given axes.

Let Plt P2, etc., be the magnitudes of the given forces ; (plt ^i),
(/2> q*)> etc., the distances of their points of application from
the central axes parallel to the two given axes ; (a, b) the dis-
tances of the centroid of the system from the given axes. Let
A and A1 be the products of inertia of the system with respect
to the given axes and the parallel central axes respectively.
Then

A =P1

But /Vi+/V2+---=o; and Pip-L + P*pz-\ — =o; since each of
these expressions is the sum of the moments of the given forces

MOMENTS OF INERTIA OF PLANE AREAS 177

with respect to an axis through the centroid of the system.

Hence,

A=A' +

which proves the proposition.

If A = (Pl + P2 + -..)c2andA'=(Pl + P2+..-y2, we have

From the above proposition it follows that if the axes have
such directions that the product of inertia with reference to the
central axes is zero, the product of inertia with reference to the
given axes is the same as if the forces all acted at the centroid.
When this condition is known to be satisfied, then for the pur-
pose of finding the product of inertia the system of forces may
be replaced by their resultant.

It follows also, in the case when the product of inertia for the
central axes is zero, that if one of the given axes coincides with
the parallel central axis, the product of inertia for the given
axes is zero ; for in this case either a or b is zero, and hence
ab(Pl + P2-\ — ) is zero. Therefore,

If the product of inertia of a system is zero for two axes, A'
and A", one of which (as A') contains the centroid of a system,
then the product of inertia is also zero for A' and any axis
parallel to A".

§ 2. Moments of Inertia of Plane Areas.

198. Elementary Areas Treated as Forces. — If any area be
divided into small elements, and a force be applied at the
centroid of each element numerically equal to its area, the
moment of inertia of this system of forces will be approximately
equal to that of the given area. The approximation will be
closer the smaller the elementary areas are taken. If the ele-
ments be made smaller and smaller, so that the area of each
approaches zero as a limit, the moment of inertia of the sup-
posed system of forces approaches as a limit the true value of
the moment of inertia of the given area.

178 GRAPHIC STATICS.

It is seen, then, that most of the general principles which
have been stated regarding moments of inertia of systems of
forces are equally applicable to moments of inertia of areas.
The practical application of these principles, however, and
especially the graphic constructions based upon them, are less
simple in the case of areas than of systems of forces such as
those already treated. The reason for this is that the system
of forces which may be conceived to replace the elements of
area consists of an infinite number of infinitely small forces,
with which the graphic methods thus far discussed cannot
readily deal. Problems of this class are most easily treated by
means of the integral calculus, especially when the areas dealt
with are in the form of geometrical figures. It is possible,
however, by graphic methods to determine approximately the
moment of inertia of any plane area ; and in many cases exact
graphic solutions of such problems are not difficult. The proof
of these methods is often most easily effected algebraically.

199. Moments of Inertia of Geometrical Figures. — The appli-
cation of the integral calculus to the determination of moments
of inertia will not be here treated. But the values of the
moments of inertia of some of the common geometrical figures
are of such frequent use that the more important of them will
be given for future reference. The moment of inertia is in each
case taken with respect to a central axis, and will be repre-
sented by /, while the radius of gyration will be called k.

Rectangle. — Let b and d be the sides, the axis being parallel
to the side b. Then

12 12

Triangle. — Let b and d be the base and altitude. Then for
an axis parallel to the base,

MOMENTS OF INERTIA OF PLANE AREAS. 179

^'2

For an axis through the vertex, bisecting the base, £2 = — ,

where b' is the projection of the base on a line perpendicular to
the axis.

Circle. — Let d be the diameter. Then

r_. 7,2 __

— /- ' — /-'

64 16

For a central axis perpendicular to the plane of the circle,

T . 7,2

1 = - > ^ ~~ IT"

32 8

Ellipse. — Let a and ^ be the semi-axes. Then for an axis
parallel to a,

4
For an axis parallel to b,

4 4

For a central axis perpendicular to the plane of the ellipse,

A2 =

Graphic construction for radius of gyration. — Whenever £2
can be expressed as the product of two known factors, the value
of k can be found by the construction already used in Art. 191.

jcy

Thus, in case of a rectangle, for which /£2 = — , we may put
d .f in pi 6 we take LM=d MN=-, the

34 43

construction there shown will give MP as the value of k. For

the triangle, the axis being parallel to the base, we have
£2 = _ . -, and the same construction is applicable. For the

36 V V

axis through the vertex bisecting the base, /£2 = — • — •

4 6

I So GRAPHIC STATICS.

200. Product of Inertia. — General principles. — Products of
inertia of areas are determined by means of the integral calculus
in a manner similar to that employed for moments of inertia.
The following fundamental principles regarding products of
inertia of geometrical figures will be found useful :

(1) With reference to two rectangular axes, one of which is
an axis of symmetry (Art. 179), the product of inertia is zero.
For it is manifest that the products of inertia of two equal
elements, symmetrically placed with reference to one of the
axes, are numerically equal but of opposite sign. Hence, if
the whole area can be made up of such pairs of elements, the
total product of inertia is zero.

(2) If the two axes are not rectangular, but the area can be
divided into elements such that for every element whose dis-
tances from the axes are /, q, there is an equal element whose
distances are /, — q, or — p, q, the product of inertia is zero.
This includes the preceding as a special case.

201. Products of Inertia of Geometrical Figures. — In each of
the following cases the product of inertia is zero :

A triangle, one axis containing the vertex and the middle
point of the base, the other being any line parallel to the base.

A parallelogram, the axes being parallel to the sides and one
axis being central. This includes the rectangle as a special
case.

An ellipse, the axes being parallel to a pair of conjugate
diameters, and one axis being central. This includes, as a
special case, that in which one axis is a principal diameter and
the other is any line perpendicular to it ; and under this case
falls also the circle.

202. Approximate Method for Finding Moment of Inertia of
Any Area. — To apply the method of Art. 190 to the determina-
tion of the moment of inertia of a plane area, we should strictly
need to replace the area by an infinite number of parallel forces,
proportional to the infinitesimal elements of the given area, and

MOMENTS OF INERTIA OF PLANE AREAS.

181

with points of application in these elements. If, instead, we
divide the given figure into finite portions whose several areas
are known, and assume forces proportional to those areas to
act at their centroids, we may get an approximate value for the
moment of inertia, which will be more nearly correct the smaller
the elements. This will be illustrated by the area shown in
Fig. 67.

Let QR be the axis with reference to which the moment of
inertia is to be found, in this case taken as a central axis.

Fig. 67

Divide the figure into four rectangular areas as shown, and
assume forces numerically equal to these areas to act at their
centroids parallel to QR. The force polygon is ABODE.
Draw the funicular polygon corresponding to a pole O, and let
the successive strings intersect QR in A', B1 ', Cr, D' , E'. Take
this as a new force polygon, and with any convenient pole
distance draw a second funicular polygon, using the same lines
of action. Let the first and last strings intersect QR in A"
and E". Then A"E'f multiplied by the product of the two
pole distances gives the moment of inertia of the four assumed
forces, and approximately the moment of inertia of the given

1 82 GRAPHIC STATICS.

figure. If the first pole distance is taken equal to AE (as is
the case in Fig. 67), the radius of gyration may be found by the
construction of Art. 191. Thus in Fig. 67, MP is the radius
of gyration as determined by this method.

A more accurate result may be reached by dividing the area
into narrower strips by lines parallel to QR ; since the narrower
such a strip is, the more nearly will the distance of each small
element from the axis coincide with that of the centroid of the
strip. If the partial areas are taken as narrow strips of equal
width, the forces may be taken proportional to the average
lengths of the several strips.

203. Accurate Methods. — If the given figure can be divided
into parts, such that the area of each is known, and also its
radius of gyration with respect to its central axis parallel to the
given axis, the above method may be so modified as to give an
accurate result. Two methods will be noticed.

(i) When the axis is known at the start. — Let the line of
action of the force representing any partial area be taken at a
distance from the given axis equal to the radius of gyration of
that area with reference to the axis. If this is done, it is
evident that the moment of inertia of the system of forces is
identical with that of the given area. When the axis is known,
the position of the line of action for any force may be found as
follows : Let QR (Fig. 68) be the given axis,
and Q'R' a parallel axis through the centroid
of any partial area. Draw KL perpendicu-
K- lar to QR, and lay off LM equal to the
radius of gyration of the partial area with

iTig.es respect to Q'R'. Then KM is the length
of the radius of gyration with respect to QR. (Art. 196.) Take
KN equal to KM and draw Q" R" through N parallel to Q'R' ;
then Q"R" is to be taken as the line of action of the force
representing the partial area in question.

(2) When the axis is at first unknown. — The method to be

MOMENTS OF INERTIA OF PLANE AREAS.

183

employed in this case is to let the force representing any partial
area act in a line through the centroid of that area ; and then
assume the force representing its moment to act in such a line
that the moment of this second force shall be numerically
equal to the moment of inertia of the partial area. This line
may be found as follows : Let k represent the radius of gyra-
tion of the partial area with respect to its central axis parallel
to the given axis, and a the distance between the two axes.
Then the moment of inertia of the partial area with respect

to the given axis is A (#2-j-/£2), if A represents the area. But

/ ^2\
A (a2l-\-k^}=Aa(a-\- — \ Hence, if a force numerically equal to

\ a)

A is assumed to act with an arm a, then a force equal to its

£2 .

moment Aa must act with an arm a-\ — in order that its moment

a
may equal A

/
The distance a-\ — can be found by a simple construction.

Let QR and Q'R' (Fig. 69) be the given axis and the central
axis respectively. Draw KL
perpendicular to QR and lay off
L M equal to k. From M draw
a line perpendicular to KM, in-
tersecting KL produced at N.

Then KN=a + k-- For> in the
a

similar triangles KNM, KML, we have
KN

KM KL

or KN=

But KL = a, and

hence

This second method is more useful than the first, because in
applying it the first funicular polygon can be drawn before the
position of the inertia-axis is known. Thus, a very common

1 84

GRAPHIC STATICS.

case is that in which the moment of inertia of an area is to be
found for a central axis, whose direction is known, while at the
outset its position is unknown because the centroid of the area
is unknown. If the second method be employed, the first
funicular polygon can be drawn at once, and serves to locate
the required central axis, as well as to determine the moments
of the first set of forces as soon as the axis is known. The
central axis and the moment of inertia with respect to it are
thus determined by a single construction.

Example. — The method last described is illustrated in Fig.
70. The area shown consists of two rectangles, the centroids
of which are marked ab and be. The moment of inertia is to
be found for a central axis parallel to the longer side of the
rectangle ab.

Fig. 70

We draw through ab and be lines parallel to the assumed
direction of the axis, and take these for the lines of action of
forces AB, BC, proportional to the areas of the two rectangles.
ABC is the force polygon, and the pole distance is taken equal

MOMENTS OF INERTIA OF PLANE AREAS. 185

to AC. The intersection of the strings 0a, oc, determines a
point of the required central axis ac. The moments of the
given forces are proportional to A'JS', B'C. We now have to
find the lines of action for the forces AB', B'C, in accordance
with the method above described.

Take any line perpendicular to the axis ac, as VQ, the side
of one of the rectangles. From R, the point in which this line
intersects the vertical line through be, lay off R T equal to the
central radius of gyration of the rectangle whose centroid is
be. To find this central radius of gyration, we know that its

value is -y— (Art. 199), where d is the length of the side of
the rectangle perpendicular to the axis. Hence we take

QR = ~, RS=—y and make QS the diameter of a semicircle,

3 4

intersecting the vertical line be in T; then R T is the required
radius of gyration of the rectangle with respect to a central
axis. Now draw from T a line perpendicular to VTt intersect-
ing VQ in U\ then the line b'c' drawn through U parallel to the
given axis is the line of action of the force B'C'.

By a similar construction applied to the other rectangle, a'&'
is located as the line of action of the force A'B'. The second
funicular polygon is now drawn, and the points A", C" are
found by the intersection of the strings o'a', o'c* with the axis.
Hence the moment of inertia of 'the given area is equal to
A"C" xHxH'. In the figure H is made equal to AC, hence
the radius of gyration can be determined by the usual con-
struction, and its length is found to be MP.

204. Moment of Inertia of Area Determined from Area of
Funicular Polygon. — The method given in Art. 193 for finding
the moment of inertia of a system of forces by means of the
area of the funicular polygon may be applied with approximate
results to the case of a plane figure. If the forces are taken
as acting at the centroids of the areas they represent, then to

1 86 GRAPHIC STATICS.

get good results these partial areas should be taken as narrow
strips between lines parallel to the axis.

If the lines of action are determined as in the first method
of the preceding article, then the area enclosed by the funic-
ular polygon and the axis represents accurately the required
moment of inertia.

205. Product of Inertia Determined Graphically. — To deter-
mine the product of inertia of any area, let it be divided into
small known parts, and let parallel forces numerically equal to
the partial areas be assumed to act at the centroids of these
parts. The product of inertia of these forces may then be
found as in Art. 194, and its value will represent approximately
the product of inertia of the given area.

If the partial areas can be so taken that the product of
inertia of each with reference to axes through its centroid par-
allel to the given axes is zero, the method here given is exact.
(Art. 197.)

If the partial areas are taken as narrow strips parallel to one
of the axes, the condition just mentioned will be nearly fulfilled ;
for the product of inertia of each strip for a pair of axes
through its centroid, one of which is parallel to its length, will
be very small.

CHAPTER X. CURVES OF INERTIA.

§ I. General Principles.

206. Relation between Moments of Inertia for Different Axes
through the Same Point. — The moment of inertia with respect
to any axis through a given point can be expressed in terms of
the moments and product of inertia for any two axes through
that point. It is necessary here to use algebraic methods, but
the results reached form the basis of graphic constructions.

Let OX, OY (Fig. 71) be the two given axes; 0, the angle
included between them ; P1} P2, etc., the forces of the system ; x\
y' , the coordinates of the point
of application of any force P,
referred to the axes OX, OY\
p, q, the perpendicular distances
of the same point from O Y and
OX respectively, so that

Let afy b1 y and c! be quantities defined by the equations

187

1 88 GRAPHIC STATICS.

Then

and a' sin 6, b' sin 0, c' sin 0 are respectively the radius of
gyration with respect to O Y, the radius of gyration with respect
to OX, and the product-radius (Art. 189) with respect to OX
and OY.

The moment of inertia (/) and radius of gyration (k) of the
system for the axis OM, making an angle <£ with OX, may now
be computed as follows :

Let s = perpendicular distance of the point of application of
any force P from OM. Then from the geometry of the figure
it is seen that

s=y' sin (9 — <$>)—xf sin <£.
Hence

/= %Ps*= 2/y2 sin2 (6 -(/>)- 2. 2/V/ sin (0 - <f>) sin 0

or,

sin2 (0-<f>) -2 c'^P - sin (0-0) sin

the factors involving 0 and 0 being constant for all terms of
the summation. Hence

^2=-^- = ^2sin2(i9-0)-2^2sin((9-(/))sin0 + ^'2sin20 . . (i)

From these equations / and k may be computed if a', b!, and
c1 are known ; that is, if the moments and product of inertia for
the two axes OX and O Fare known.

Special case. — If # = 90°, the equation (i) becomes

£W2cos20-2<:'2sin</> cos 0 + 0'2 sin2c/> ........ (2)

GENERAL PRINCIPLES. 189

207. Products of Inertia for Different Axes through the Ori-
gin.— The product of inertia with respect to OM and OX may
be found as follows :

Let A — the required product of inertia ; then

A = ^Pqs = ^Pyf sin0 [/ sin (0-0)-*' sin 0]
= 2P [/2 sin 0 sin (0-0) -*'/ sin 9 sin 0]
= [£'2 sin 6 sin ((9 - 0) - ^'2 sin 0 sin 0] 2P.
Let // = product-radius for axes OM and OX; then

//2= A = £'2 sin 0 sin (0-0) -V2 sin 0 sin 0.

Special case. — The axis OM may be so chosen that A = o.
This will be the case if

£'2sin(0-0)=<:'2sin0.

208. Inertia Curve. — If on OM (Fig. 71) a point M be taken
such that the length OM depends in some given way upon the
value of k, and if similar points be located for all possible direc-
tions of OM, the locus of such points will be a curve which is
called a curve of inertia of the system for the center O.

The form of the curve will depend upon the assumed law
connecting O M with k.

209. Ellipse or Hyperbola of Inertia. — The simplest curve is

obtained by assuming OM to be inversely proportional to k.

cfi sin2 0
Let OM=r, and take r2 = — > where d* is a positive

quantity, so that d always represents a real length, positive or
negative.

Equation (i) of Art. 206 then becomes

d^ sin2 0 = #'Va sin2 (0 — 0) — 2 c'^r* sin (0 — 0) sin 0 + tf'V2 sin20,

which is the polar equation of the inertia-curve, r and 0 being
the variable coordinates. Let x, y be coordinates of the point
M referred to the axes OX, O Y. Then

r _ x y

sin 0 sin (0 — 0) sin 0

190 GRAPHIC STATICS.

r2 sin2 (0 — 6)

whence — ^L=x\

sin20

r2 sin2 c)

sin20 '

* sin 0 — () sin

and the equation becomes

J* = l,lW-2c'2xy + aiy ........... (3)

The form of this equation shows that it represents a conic
section whose center is at the origin of coordinates O. This
conic may be either an ellipse or a hyperbola.

The equation will be discussed more fully in a later article.
One fact may, however, be here noticed.

If the moment of inertia / is positive for all positions of the
axis, the radius of gyration k will be real, whatever the value
of (f). But r, the radius vector of the curve, will be real when k
is real ; hence in this case the curve is an ellipse. This is
always the case if the given forces have all the same sign.

If the forces have not all the same sign, it is possible that
the value of / may have different signs for different directions
of the axis. If this is so, certain values of $ make k (and
therefore r} imaginary. In this case the curve is a hyperbola.

The most important case is that in which the given forces
have all the same sign which may be taken as plus, so that the
moment of inertia is always positive, and the curve an ellipse ;
and to this case the discussion will be confined.

§ 2. Inertia-Ellipses for Systems of Forces.

210. Properties of the Ellipse. — In discussing ellipses of in-
ertia use will be made of certain general properties of the
ellipse, which, for convenience of reference, will be here sum-
marized. For the proof of the propositions stated the reader
is referred to works on the conic sections.

INERTIA-ELLIPSES FOR SYSTEMS OF FORCES.
(i) The equation

represents an ellipse if F2 — AC is negative; a hyperbola if
B2 — AC is positive. (Salmon's Conic Sections, p. 140.) The
coordinate axes may be either rectangular or oblique.

(2) Two diameters of an ellipse are said to be conjugate to
each other if each bisects all chords parallel to the other. If
the axes of coordinates coincide with a pair of conjugate
diameters, the lengths of which are 2 a' and 2 b' , the equation
of the curve is

A particular case of this equation is that in which the coordi-
nate axes are rectangular, being the principal axes of the curve ;
in which case we may write a and b instead of a' and b' .

(3) In an ellipse, the product of any semi-diameter and the
perpendicular from the center on the tangent parallel to that
semi-diameter is constant and equal to ab. That is, if r is any
radius vector of the curve drawn from the center, and / the
length of the perpendicular from the center to the parallel
tangent, we have

pr=ab
where a and b are the principal semi-axes of the curve.

(4) Let a' and b1 be conjugate semi-diameters. Then each is
parallel to the tangent at the extremity of the other. Hence
the length of the perpendicular from the center to the tangent
parallel to a1 is b' sin 0, where 6 is the angle included between
a' and b1 '. Therefore from the preceding paragraph,

afb' sin 6 = ab.

(5) An ellipse can be constructed, when a pair of conjugate
diameters is known, as follows :

IQ2 GRAPHIC STATICS.

Let AAf, BB! (Fig. 72), be the conjugate diameters, O being
the center of the ellipse. Complete the parallelogram OBCA.

Divide OA and CA into parts
proportional to each other, be-
ginning at O and C. Through
the points of division of OA
draw lines radiating from B\
and through the points of divi-
sion of CA draw lines radiating
from B. The points of inter-
section of the corresponding lines in the two sets are points of
the ellipse. In a similar way, the other three quadrants may
be drawn. (The location of one point is shown in the figure.)

A convenient way to locate the corresponding points of
division on OA and CA is to cut-these lines by lines parallel to
the diagonal OC.

211. Discussion of Equation of Inertia-Curve. — We will now
examine the equation of the inertia-curve,

with reference to the properties of the ellipse above enumerated.

(i) If a^b^ — c^ is positive, the equation denotes an ellipse.
This cannot be the case if a'2 and b'* have opposite signs. But
from the definitions of a<2> and b^ (Art. 206) it is seen that their
signs are the same as those of the moments of inertia for Kand
X axes respectively. Hence, if there are any two axes through
the assumed center for which the moments of inertia have
opposite signs, the inertia-curve is a hyperbola.

If the moment of inertia has the same sign for all axes
through the assumed center, the curve is an ellipse. For, since

c'2 sin20 = PV (Art. 206), c' may be made zero by choosing the
^x

axes so that the product of inertia with respect to them is
zero ; and if c1 is zero, and an and b^ have the same sign, the
quantity a^b^ — c* is positive.

INERTIA-ELLIPSES FOR SYSTEMS OF FORCES. 193

This agrees with the conclusion stated in Art. 209.
We shall here deal only with ellipses of inertia.

(2) If c' = o, the coordinate axes are conjugate axes of the
curve. But the condition c' = o means that the product of
inertia for the two axes is zero. Hence any two axes for which
the product of inertia is zero are conjugate axes of the inertia-
curve. (This is true whether the curve is an ellipse or a
hyperbola.)

(3) By the law of formation of the inertia-conic (Art. 209),
the length of the radius vector lying in any line is inversely
proportional to the radius of gyration with respect to that line.
But by (3) of the last article, the perpendicular from the center
on the tangent parallel to any radius vector is inversely propor-
tional to the length of that radius vector. Hence the perpen-
dicular distance between any diameter and the parallel tangent
is directly proportional to the radius of gyration with respect to
that diameter. The curve may be so constructed that the
length of this perpendicular is equal to the radius of gyration,
as follows :

From Art. 209, we have

, ^2sin#

r
and from (3) and (4) of the last article we have

ab a'b' sin 6
/ = — = — — ,
r r

if a' and V are conjugate semi-diameters. Now take

a"2 sin 0 = a& = a'&' sin 0,
or dz=a'b',

and we have >£=/>

and the equation of the curve becomes (since c' = o when the
axes are conjugate)

194 GRAPHIC STATICS.

If the equation be written in this form, a' and b1 having the
meanings assigned in Art. 206, the radius of gyration about any
axis througJi tJie center of the ellipse is equal to tJie perpendictdar
distance between the axis and the parallel tangent to the ellipse.

Hereafter we shall mean by inertia-ellipse the curve obtained
by taking dz=a'b' as above described, so that the radius of gyra-
tion for any axis can be found by direct measurement when a
parallel tangent to the ellipse is known.

212. To Determine Tangents to the Inertia-Ellipse for Any
Center. — Let the radius of gyration (k) be found for any
assumed axis through the given center by one of the methods
already described (Arts. 202 and 203). Then two lines par-
allel to the axis and distant k from it, on opposite sides, will be
tangents to the inertia-ellipse.

213. To Construct the Inertia-Ellipse, a Pair of Conjugate
Axes Being Known in Position. — If the positions of two axes
conjugate to each other can be found, the ellipse can be drawn
by the following method :

Determine the radius of gyration for each of the two axes
and draw the corresponding tangents as in the preceding
article ; then proceed as follows :

Let XX', YY' (Fig. 73) be the given axes, and let the four

tangents determined as above form the parallelogram PQRS.

/Y- Let A, A', B, B' be the points

P ^- -y Q in which these tangents in-

/ / / tersect the axes XX1, YY'.

x' / / / x

A/ /o /A Then, since each diameter is

/ / / parallel to the tangents at the

extremities of the conjugate
'3 diameter, A, A', B, B' are the

extremities of the diameters lying in the given axes. The
ellipse can now be constructed as explained in Art. 210
(Fig, 72).

This method of constructing the inertia-ellipse is useful

INERTIA-CURVES FOR PLANE AREAS. 195

whenever the given system has a pair of conjugate axes which
can be located by inspection.

214. Central Ellipse. — It is evident that an inertia-curve
can be found with its center at any assumed point. That
ellipse whose center is the centroid of the given system is called
the central ellipse for the system.

Since the central ellipse gives at once the radius of gyration
for every axis through the centroid of the system, it enables us
to determine readily the radius of gyration for any axis what-
ever, by means of the known relation between radii of gyration
for parallel axes. (Art. 196.)

§ 3. Inertia-Curves for Plane Areas.

215. General Principles. — The principles deduced in the
treatment of inertia-curves for systems of forces are all true for
the case of plane areas. But special difficulties arise in dealing
with areas, because of the fact that the system of forces equiva-
lent to any area consists of an infinite number of forces. The
principles already developed are, however, sufficient to deal at
least approximately with all areas, and accurately with many
cases.

216. Inertia-Curve an Ellipse. — Since the forces conceived
to replace the elements of area (Art. 198) have all the same
sign, the value of k* is always positive, and the inertia-curve is
always an ellipse. (Arts. 209 and 211.)

2 1 7. Cases Admitting Simple Treatment. — Whenever a pair
of conjugate diameters can be located, and the radius of gyra-
tion determined for each, the inertia-ellipse can be at once
drawn as in Art. 213. This will be the case whenever it is
possible to locate readily a pair of axes for which the product of
inertia is zero.

(i) If there is an axis of symmetry, this and any line perpen-
dicular to it are a pair of conjugate axes (and in fact the principal
axes) of the inertia-ellipse whose center is at their intersection.
(Art. 200.)

GRAPHIC STATICS.

(2) If two axes can be located in such a way that for every
element of area whose distances from the axes are /, q, there is
an equal element whose distances are /, —qy or — /, q, the
product of inertia is zero for the two axes, and these are there-
fore a pair of conjugate axes of the inertia-ellipse whose center
is at their intersection. This of course includes the case when
there is an axis of symmetry. (Art. 200.)

When a pair of conjugate axes is known, the radius of gyration
is to be found by one of the methods of Art. 202 or Art. 203 ;
the ellipse can then be drawn exactly as explained in Art. 213.

If a pair of conjugate axes cannot be located by inspection,
the inertia-ellipse cannot be so readily constructed. Such cases
will not be here treated.

As examples of areas, in which the principal axes of the
inertia-curve can be located by inspection, may be mentioned
the cross-section of the I-beam, the deck-beam, the channel-bar,
and other shapes of structural iron and steel.

Many geometrical figures possess axes of symmetry. In
others a pair of conjugate axes can be located by principle (2).
Some of these will be discussed in the next article.

Example. — Draw the central ellipse for the deck-beam sec-
tion shown in Fig. 74.
^

[SUGGESTIONS. — Since there is an axis of symmetry, this
contains one of the principal axes of the ellipse. The other
can be drawn as soon as the centroid of the section is
known. Find the radius of gyration for each axis by
Art. 202, and then construct the ellipse as explained in
Art. 213.]

2 1 8. Central Ellipses for Geometrical Fig-
ures. — In many of the simple geometrical
figures, not only can a pair of conjugate
axes be located by inspection, but the radius
of gyration for each of these axes can be found by a simple
construction, so that the central ellipse can be readily drawn.
Some of these cases will be here summarized.

INERTIA-CURVES FOR PLANE AREAS. 197

(i) Parallelogram.— Let ABCD (Fig. 75) be the parallelo-
gram; then XX', YYf, drawn through the centroid parallel to
the sides, are a pair of
conjugate axes of the cen-
tral ellipse. Let AB = b,
BC=d, and let //= the
perpendicular distance be-
tween AB and DC. The
moment of inertia of the
parallelogram with re-
spect to the axis XX' is
equal to the moment of inertia of a rectangle of sides b and h.
Hence $, the square of the radius of gyration for this axis, is
— • The length of k can be found by the construction used in

case of the rectangles in Fig. 70. The following modification
of the method is, however, more convenient :

Make EF=\BC, EG = \BC, and draw a semicircle with FG
as a diameter. From E draw a line perpendicular to BC, inter-
secting the semicircle at /. Lay off EH=EI\ then a line
through H parallel to XX1 is a tangent to the central ellipse.
For by construction,

V 12

And since the projection of BC on a line perpendicular to XX is
equal to Ji, the projection of EH on. the same line is equal to — l—,

•\/I2

that is to k.

The tangent parallel to the side BC may be found in a simi-
lar way. It may, however, be located more simply as follows :
It is evident that the distance between YY' and a tangent par-
allel to it, measured along AB, bears the same ratio to AB that
EH does to BC. Hence, the parallelogram formed by the four
tangents, two parallel to XX' and two parallel to YY't is simi-
lar to the parallelogram ABCD.

$ GRAPHIC STATICS.

Fig. 75 shows this parallelogram and also the ellipse.

(2) Triangle. — Let ABC (Fig. 76) be the triangle; b =
,Y length of the base BC\ b1 =

length of projection of BC on a
line perpendicular to AD ; ^ = al-
titude measured perpendicular to
-#£ ^ and a line through the
centroid parallel to BC are a pair
of conjugate axes of the central
ellipse (Art. 217). From Art.
199, the radius of gyration for a
central axis parallel to BC is

\— = -\/r • f- Let XX' be this axis, and H the point in which

it intersects AC. Then HC=\AC. Take HK=\AC=\HC,
and make ^£7 the diameter of a semicircle. From //"draw ///
perpendicular to AC, intersecting the semicircle at /. Make
HL = HI\ then the line through L parallel to XX' is a tangent
to the central ellipse. For the radius of gyration with respect
to XX' is to HL as the altitude ^is to AC.

Again, for the axis AD, the radius of gyration is ^/ — (Art.

199). Make DE=% BC and DF=\ BC, and take EF as a
diameter of a semicircle. From D draw a line perpendicular to
BC, intersecting the semicircle in M\ and make DG=DM; then
a line from G parallel to AD is a tangent to the central ellipse.
The figure shows the parallelogram formed by the two tan-
gents parallel to XX' and the two parallel to AD, and also the
central ellipse.

(3) Ellipse. — From Art. 199, the radii of gyration of an
ellipse with respect to the two principal diameters are \a
and \ b. Hence the central ellipse of inertia is similar to the
given ellipse, its semi-axes being -|- a and \ b. A special case
of this is a circle, for which the central curve is a circle whose
radius is half that of the given circle.

INERTIA-CURVES FOR PLANE AREAS.

199

(4) Semicircle. — Let ABC (Fig. 77) be the semicircle, O
being the centroid. (The point O may be located by the
method described in Art. 184.)
From symmetry it is evident
that the principal axes of the
central ellipse are XX' and
YY', drawn through O, re-
spectively parallel and per-
pendicular to AB.

With respect to the axis
YY't the radius of gyration
is evidently -|- r, the same as

for the whole circle. Hence two lines parallel to YY' and dis-
tant J r from it are tangents to the central ellipse.

Again, the radius of gyration of the semicircle with respect
to AB as an axis is also -|- r, the same as for the whole circle.
To find it for the axis XX' , with D as a center, and radius ^ r,
draw an arc intersecting XX1 at F\ then OF2 — DF2 — OD2.
But DFis equal to the radius of gyration with respect to AB,.
and OD is the distance between XX' and AB ; hence (Art. 196)
OF is equal to the radius of gyration with respect to XX'.
Hence if two lines are drawn parallel to XX' , each at a distance
from it equal to OF, they will be tangents to the central ellipse.
The ellipse can now be drawn in the usual manner.

219. Summary of Results. — By the principles and methods
'developed in the present chapter, inertia-curves can be drawn
for all the simpler cases that may arise ; namely, whenever a
pair of conjugate axes can be located by inspection. This will
be the case whenever the product of inertia can be seen to be
zero for any pair of axes ; and it includes every case of an area
possessing an axis of symmetry.

It is believed that this chapter contains as complete a discus-
sion as is needed by the student of engineering. Those who
desire to pursue the subject further may consult other works.

Fig. 34

ScaZe, iinch=6feet.

(B)

Scate, I incU=1800 Ibs.

PLATE I.

DEF

Scale, 1 inch =6, ooo Ibs.
(0

Scale: I in,= 4,000 Ibs,

Scale: I in

Fig". 42

PLATE II.

DEFG

(E)
Scale: I in, =5,OOO Ibs.

8.1

5.S

o O OOO C

Linear Scale, 1 in. = 20 ft.
Force Scale, 1 in. = 80,000 Tbs.

(S)

PLATE III.

II I] i i| ll II il
s id s H - - o-

; i.5 1.5 T.I 1.8 | 5.T [ 1.8 [ l.C

o o o O OOO o o o o [

1500 Ibs. per foot

PLATE IV!

(A)

PLATE V.

inear Scale, 1 in.- 20 ft.
jrce Scale, 1 in.- 80 ,000 Ibs.

Linear Scab
JFarce Scale,

PLATE VI,

In.- 20 ft.
,= 80,000 Ibs.

PLATE VII.

Linear Scale, 1 w.- 60ft.
Force Scale,! in.- 120,000 Ibs,

Fig. 48

(A)

* --*--

^-' — — D.

PLATE VIII.

THE ELEMENTS OF PHYSICS.

EDWARD L. NICHOLS, B.S., Ph.D.,

Professor of Physics in Cornell University,

AND

WILLIAM S. FRANKLIN, M.S.,

Professor of Physics and Electrical Engineering at the Iowa Agricultural College, Ames, fa,

WITH NUMEROUS ILLUSTRATIONS.

fVol. I. Mechanics and Heat.

PART I. In Three Volumes : -< II. Electricity and Magnetism.

( III. Sound and Light.

Price $1.50, net, per volume.

It has been written with a view to providing a text-book which shall correspond with
the increasing strength of the mathematical teaching in our university classes. In most of
the existing text-books it appears to have been assumed that the student possesses so
scanty a mathematical knowledge that he cannot understand the natural language of
physics, i.e., the language of the calculus. Some authors, on the other hand, have assumed
a degree of mathematical training such that their work is unreadable for nearly all under-
graduates.

The present writers having had occasion to teach large classes, the members of which
were acquainted with the elementary principles of the calculus, have sorely felt the need of
a text-book adapted to their students. The present work is an attempt on their part to
supply this want. It is believed that in very many institutions a similar condition of affairs
exists, and that there is a demand for a work of a grade intermediate between that of the
existing elementary texts and the advanced manuals of physics.

No attempt has been made in this work to produce a complete manual or compendium
of experimental physics. The book is planned to be used in connection with illustrated
lectures, in the course of which the phenomena are demonstrated and described. The
authors have accordingly confined themselves to a statement of principles, leaving the
lecturer to bring to notice the phenomena based upon them. In stating these principles,
free use has been made of the calculus, but no demand has been made upon the student
beyond that supplied by the ordinary elementary college courses on this subject.

Certain parts of physics contain real and unavoidable difficulties. These have not been
slurred over, nor have those portions of the subject which contain them been omitted. It
has been thought more serviceable to the student and to the teacher who may have occa-
sion to use the book to face such difficulties frankly, reducing the statements involving
them to the simplest form which is compatible with accuracy.

In a word, the Elements of Physics is a book which has been written for use in such
institutions as give their undergraduates a reasonably good mathematical training. It is
intended for teachers who desire to treat their subject as an exact science, and who are
prepared to supplement the brief subject-matter of the text by demonstration, illustration,
and discussion drawn from the fund of their own knowledge.

THE MACMILLAN COMPANY.

NEW YORK: CHICAGO:

66 FIFTH AVENUE. ROOM 23, AUDITORIUM.

A Laboratory flanual

Physics and Applied Electricity.

ARRANGED AND EDITED BY

EDWARD L. NICHOLS,

Professor of Physics in Cornell University.
IN TWO VOLUMES.

VoL L JUNIOR COURSE IN GENERAL PHYSICS.

ERNEST MERRITT and FREDERICK J. ROGERS.
Cloth. $3.00.

Vol. II. SENIOR COURSES AND OUTLINE OF
ADVANCED WORK.

GEORGE S. MOLER, FREDERICK BEDELL, HOMER J. HOTCHKISS,
CHARLES P. MATTHEWS, and THE EDITOR.

Cloth, pp. 444. $3.25.

The first volume, intended for beginners, affords explicit directions adapted to a modern
laboratory, together with demonstrations and elementary statements of principles. It is
assumed that the student possesses some knowledge of analytical geometry and of the cal-
culus. In the second volume more is left to the individual effort and to the maturer intel-
ligence of the practicant.

A large proportion of the students for whom primarily this Manual is intended, are pre-
paring to become engineers, and especial attention has been devoted to the needs of that
class of readers. In Vol. II., especially, a considerable amount of work in applied elec-
tricity, in photometry, and in heat has been introduced.

COMMENTS.

" The work as a whole cannot be too highly commended. Its brief outlines of the
various experiments are very satisfactory, its descriptions of apparatus are excellent; its
numerous suggestions are calculated to develop the thinking and reasoning powers of the
student. The diagrams are carefully prepared, and its frequent citations of original
sources of information are of the greatest value." — Street Railway Journal.

" The work is clearly and concisely written, the fact that it is edited by Professor Nichols
being a sufficient guarantee of merit." — Electrical Engineering,

" It will be a great aid to students. The notes of experiments and problems reveal
much original work, and the book will be sure to commend itself to instructors."

r— San Francisco Chronicle.

THE MACMILLAN COMPANY,

NEW YORK: CHICAGO:

66 FIFTH AVENUE. ROOM 23, AUDITORIUM.

A LABORATORY MANUAL

PHYSICS AND APPLIED ELECTRICITY.

ARRANGED AND EDITED BY

EDWARD L. NICHOLS.

COMMENTS.

The work as a whole cannot be too highly commended. Its brief outlines of the
various experiments are very satisfactory, its descriptions of apparatus are excellent ;
its numerous suggestions are calculated to develop the thinking and reasoning powers
of the student. The diagrams are carefully prepared, and its frequent citations of
original sources of information are of the greatest value. — Street Railway Journal.

The work is clearly and concisely written, the fact that it is edited by Professor
Nichols being a sufficient guarantee of merit. — Electrical Engineering.

It will be a great aid to students. The notes of experiments and problems reveal
much original work, and the book will be sure to commend itself to instructors.

— S. F. Chronicle.

Immediately upon its publication, NICHOLS' LABORATORY MANUAL OF
PHYSICS AND APPLIED ELECTRICITY became the required text-book in
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versity ; University of Nebraska ; Brooklyn Polytechnic Institute ;
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College ; Mt. Holyoke College ; etc., etc.

It is used as a reference manual in many other colleges where the
arrangement of the courses in Physics does not permit its formal
introduction.

THE MACMILLAN COMPANY.

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WORKS ON PHYSICS.

A TEXT-BOOK OF THE PRINCIPLES OF PHYSICS.

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A LABORATORY MANUAL OF EXPERIMENTAL PHYSICS.

By W. J. LOUDON AND J. C. flcLENNAN,

Demonstrators in Physics, University of Toronto.
Cloth. 8vo. pp. 302. Price, $1.90, net.

The book contains a series of elementary experiments specially adapted for students who have had
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WORKS BY R. T. GLAZEBROOK, M.A.

^MECHANICS AND HYDROSTATICS. Containing: I. Dynamics. II. Statics. III. Hydro-
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.STATICS, pp. 180. Price, 90 cents, net. DYNAMICS, pp. 256. Price, $1.25, net.

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BY S. L. LONEY,

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ELEMENTS OF STATICS AND DYNAMICS.
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MECHANICS AND HYDROSTATICS FOR BEGINNERS.
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